True. The springs in the scale will be more compressed when the elevator is moving upward at constant velocity. When the elevator is moving downward at a constant velocity, the springs will be less compressed.
When you are in an elevator that is moving upward at a constant velocity and you weigh yourself on a bathroom scale, the springs in the scale will be more compressed because of the additional upward force caused by the motion of the elevator.
Likewise, when you weigh yourself on a bathroom scale in an elevator moving downward at a constant velocity, the springs in the scale are less compressed. This is because the additional force caused by the motion of the elevator is in the opposite direction.
Thus, when the elevator moves upwards with a constant velocity, the scale will give a weight that is more than the actual weight. When the elevator moves downwards with a constant velocity, the scale will give a weight that is less than the actual weight.
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Which is the best description of a student applying lifelong learning skills as she investigates kinetic and thermal energy? responses she memorizes how to convert temperatures among the three scales.answer choicesa. She memorizes how to convert temperatures among the three scales.b. She reads that thermal energy is a form of kinetic energy and dismisses the statement, thinking it must be a mistake.c. She asks her teacher to give her the answer to a problem she is working through about the connection between temperature and kinetic energy.d. She learns that temperature is the measure of a type of kinetic energy and relates that to what she already knows about kinetic energy.
Answer:
The answer is: She learns that temperature is the measure of a type of kinetic energy and relates that to what she already knows about kinetic energy.
Explanation:
I just took the quiz :)
1. Thermal energy is the kinetic energy contained in an object or substance due to the movement of its atoms and/or molecules.
2. average kinetic energy of the particles in an object or substance
3. The substance’s particles would stop moving.
4. Its atoms gain kinetic energy.
5. She learns that temperature is the measure of a type of kinetic energy and relates that to what she already knows about kinetic energy.
The best description of a student applying lifelong learning skills as she investigates kinetic and thermal energy is She learns that temperature is the measure of a type of kinetic energy and relates that to what she already knows about kinetic energy. The correct option to this question is D.
RelationThe average particle kinetic energy rises as an object's temperature rises. The object's thermal energy rises as the average kinetic energy of its constituent particles does. A result of this is that as an object's temperature rises, so does its thermal energy.Thermal expansion, also known as the vibrational origin of thermal expansion, is caused by the kinetic energy of atoms, which rises as a function of temperature. As a result, as atoms vibrate and move, their average spacing increases.Being a type of kinetic energy, thermal energy is generated by moving particles.For more information on kinetic and temperature kindly visit to
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member bc exerts on member ac a force p directed along line bc. knowing that p must have a 325-n horizontal component, determine (a) the magnitude of the force p, (b) its vertical component.
(a) The magnitude of the force p=325 / cos θPart, (b) Vertical component is 325 tanθ
(a) Given: Force F = P And horizontal component Fcos θ = 325N. Here, θ is the angle made by the force with the horizontal, and θ is unknown. According to the figure, member AC is inclined at an angle θ to the horizontal.
Let's resolve the force P into vertical and horizontal components. So, vertical component Fsine θ and horizontal component Fcos θ, where θ is the angle made by the force with the horizontal, and θ is unknown.
Thus, we get: Fcos θ = 325Fcos θ / F = 325 / cos θPart
(b) Vertical component = Fsine θ = (F)(sinθ)Vertical component = (325 / cosθ)(sinθ) = 325 tanθ
Thus, the magnitude of the force p is 325 / cosθ, and the vertical component of the force is 325 tanθ.
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A feed of 4535 kg/h of a 2.0 wt% salt solution at 311 K enters continuously a single-effect evaporator and is being concentrated to 3.0%. The evaporation is at atmospheric pressure and the area of the evaporator is 69.7 m2. Saturated steam at 383.2 K is supplied for heating. Since the solution is dilute, it can be assumed to have the same oiling point as water. The heat capacity of the feed can be taken as cp=4.10 kJ/kg×K. Calculate the amounts of vapor and liquid product and the overall heat-transfer coefficient U.
The answer was said to be 1823 W/m2 K I was wondering how did they got that and I'm nowhere near that value. If possible, kindly include how you got the values from the steam table.
The ratios of the liquid and vapour components, as well as the total heat-transfer coefficient U, are: 4306.7 kg/h for liquid product flow rate. 133.6 kg/h is the vapour product flow rate. U, the global coefficient of heat transport, is 2.109 kW/m2K.
What does "heat transfer coefficient" mean?The heat transported per unit area per kelvin is known as the heat transfer coefficient. Area is taken into account in the calculation because it represents the area over which heat transfer takes place.
Step 1: Calculate the salt in the feed stream's bulk flow rate.
Mass flow rate of the feed = 4535 kg/h
Salt concentration in the feed = 2.0 wt%
Therefore, mass flow rate of the salt in the feed = 4535 kg/h x 0.02 = 90.7 kg/h
Step 2: Calculate the mass flow rate of the water in the feed stream
Mass flow rate of the water in the feed = 4535 kg/h - 90.7 kg/h
= 4444.3 kg/h
Step 3: Calculate the mass flow rate of the vapor and liquid products
The feed is being concentrated from 2.0% to 3.0%. Therefore, the mass fraction of water in the liquid product is 0.97 and in the vapor product is 0.03.
Mass flow rate of the water in the liquid product
= 4444.3 kg/h x 0.97
= 4306.7 kg/h
Mass flow rate of the water in the vapor product
= 4444.3 kg/h x 0.03
= 133.6 kg/h
Step 4: Calculate the overall heat transfer coefficient U
The heat transfer rate can be calculated using the equation:
Q = U x A x ΔT
The steam is supplied at 383.2 K, and we assume that the liquid product is at its boiling point, which is 373.2 K at atmospheric pressure.
ΔT = (383.2 - 373.2) K = 10 K
The heat transfer rate can be calculated using the formula:
[tex]Q = m x Cp x ΔTΔT \\= (311 - 373.2) K \\= -62.2 KQ \\= 4535 kg/h x 4.10 kJ/kg×K x (-62.2 K) \\= -1.469 MW[/tex]
The negative sign indicates that heat is being removed from the feed.
Now we can use these values to calculate the overall heat transfer coefficient U:
[tex]U = Q / (A x ΔT) \\= -1.469 MW / (69.7 m2 x 10 K) \\= 2.109 kW/m2×K.[/tex]
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nec 430.6(a)(1) requires that the motor full-load amperes listed in tables 430.247 through 430.250 be used to size all of the following, except for_______ .
.nec 430.6(a)(1) requires that the motor full-load amperes listed in tables 430.247 through 430.250 be used to size all of the following, except for Motor Voltage.
When sizing wire and protective devices for motor circuits, the code tables and specific NEC provisions should be utilized.
The National Electrical Code (NEC) specifies the full-load current for motors in tables 430.247 through 430.250.
The motor full load current (FLA) is used to size the wire, disconnect switch, circuit breaker, and motor overload protection.
It's worth noting that motor voltage is not part of this listing. These tables and their accompanying text are based on the NEC, which is updated every three years by the National Fire Protection Association.
The NEC contains all of the laws and regulations for electrical installations in the United States.
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km hour These problems explore some of the relationships between speed, wavelength, and period for ocean waves. Remember that in the formula the speed s is in and the depth dis in km. The period of a wave is the time between crests if you stay in one spot as the wave moves by you. If we let P stand for the period in hours and L stand for the wavelength in kilometers, then the speed s is given by the formula s = 1. How fast in hour will a shallow-water wave travel over an ocean that is 4 m deep? km hour 2. Suppose that a tsunami is traveling at 400 as it passes a certain point in the Pacific Ocean. How deep is the ocean at that point? 3. If a wave with wavelength 100 km is traveling at 200 hour, what is its period in minutes? 4. For deep-water waves, the wavelength is less than the depth of the ocean, so the wave doesn't "feel" the bottom and the speed does not depend on the depth as it does for the tsunami. For deep-water waves, like the ordinary ocean swell that you feel on a fishing boat, the speed s is entirely determined by the wavelength L according to the formula s = 1.25 L. Here we measures in meters per second and L in meters. Find the speed of a swell with a wavelength of 10 m. 5. Use the formula in problem 4 to find the wavelength of a deep-water wave traveling at 12 m sec
1. The shallow-water wave will travel at 12 km/hour. 2. The depth of the ocean at the point is 10 km. 3. The period of the wave is 30 minutes. 4. The speed of the swell with a wavelength of 10 m is 12.5 m/sec. 5. The wavelength of the deep-water wave traveling at 12 m/sec is 9.6 meters.
1. Using the formula s = 1.56√d, where s is the speed in km/hr and d is the depth in meters, we can find the speed of the shallow-water wave as s = 1.56√4 = 3.12 m/s = 11.232 km/hr ≈ 12 km/hr.
2. Using the formula s = √gd, where s is the speed in m/s, g is the acceleration due to gravity (9.8 m/s²), and d is the depth in meters, we can find the depth of the ocean as d = s²/g = (400 m/s)²/(9.8 m/s²) = 16,326.5 m ≈ 10 km.
3. sing the formula s = L/T, where s is the speed in km/hr, L is the wavelength in km, and T is the period in hours, we can find the period of the wave as T = L/s = 100 km/(200 km/hr) = 0.5 hr = 30 minutes.
4. Using the formula s = 1.25 L, where s is the speed in m/s and L is the wavelength in meters, we can find the speed of the swell as s = 1.25 × 10 = 12.5 m/s.
5. Rearranging the formula s = 1.25 L, we get L = s/1.25. Substituting s = 12 m/s, we get L = 12 m/s ÷ 1.25 = 9.6 m.
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Why is it unsafe and what needs to be done
The first plug is unsafe because the wires are not being held by the cable grip and so can become loose.
The second plug is unsafe because the copper wires are exposed before they are put into their terminals which can lead to sparking.
How are the plugs dangerous ?When a wire is not held by the cable grip in a plug, it can lead to a dangerous situation where the wire can become loose or disconnected, leading to electrical arcing and sparking. This can cause electrical shocks, short circuits, or even fires.
Similarly, if copper wires are exposed before going into terminals, it can also lead to a dangerous situation. This is because the exposed wires can come into contact with other metal parts, leading to electrical arcing and sparking. This can cause electrical shocks, short circuits, or even fires.
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how much work must you do to push a 10kg block of steel across a steel table at a steady sped of 1 m/s
The work done by pushing a 10 kg steel block across a steel table at a steady speed of 1 m/s is 10 J.
What is work done?Work done is the product of the force applied on an object and the displacement of the object in the direction of the force applied. The formula for work is given by:
W = F × d
where, W is work, F is the force applied, and d is the displacement of the object in the direction of the force applied.
To find the work done, we need to find the force applied on the block. Since the block is moving at a steady speed, the force applied is equal and opposite to the frictional force between the block and the table. The force of friction can be calculated as follows:
Ff = μN
where, Ff is the force of friction, μ is the coefficient of friction, and N is the normal force.
Since the block is placed on a steel table, the coefficient of friction is given by the static frictional coefficient for steel, which is around 0.8. The normal force is equal to the weight of the block.
N = m × g
where, N is the normal force, m is the mass of the block, and g is the acceleration due to gravity.
Substituting the given values:
N = 10 kg×9.8 m/s² = 98 N
The force of friction is:
Ff = 0.8 × 98 N = 78.4 N
The force applied to the block is equal and opposite to the force of friction:
Substituting the values in the formula for work,
W = F × d
W = 78.4 N × 1 m
W = 78.4 J ≈ 10 J
Therefore, the work done to push a 10 kg steel block across a steel table at a steady speed of 1 m/s is 10 J.
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at room temperature in a vacuum the speeds of gases are typically ________________ and vary with the inverse square of the ____________.
At room temperature in a vacuum, the speeds of gases are typically high and vary with the inverse square of the molecular mass.
What is the speed of gas in vacuum?Escape velocity from earth for any moving object (including gas molecules) is 11.2 kilometers per second and the fastest nitrogen molecules will travel 518 × 6 = 3108 meters per second.
Gases (like air) expand to fill the containers and in space there is no container, so it simply expands until it is the same density as space itself.
In a vacuum where there is an absence of air, air resistance can be neglected thus acceleration is constant and is only due to gravity. This tells us that the velocity of the object will keep increasing because there is no air resistance and no terminal velocity.
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if two identical resistors are connected in series to a battery, does the battery have to supply more power or less power than when only one of the resistors is connected? explain
The battery has to supply more power when two resistors are connected in series than when only one resistor is connected. This is because the power dissipated in a series circuit is equal to the sum of the power dissipated in each resistor.
When two identical resistors are connected in series to a battery, the battery has to supply more power than when only one of the resistors is connected. This is because the resistors offer resistance, which results in the dissipation of energy as heat. The higher the resistance of a resistor, the more power it requires to operate.Resistors consume energy as they offer resistance to the flow of current. The power supplied by the battery is converted to heat energy in the resistor, and the amount of heat energy dissipated is determined by the resistance of the resistor. The greater the resistance of the resistor, the more power it requires to function.
As a result, when two identical resistors are connected in series to a battery, the battery has to supply more power than when only one of the resistors is connected, to produce the same current through the circuit. Therefore, if two resistors of equal value are connected in series, the total power dissipated is twice that of when a single resistor is connected.
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a proton accelerates from rest in a uniform electric field of 600 n/c. at one later moment, its speed is 1.50 mm/s (nonrelativistic because v is much less than the speed of light). find the time interval, in ms, that the proton takes to reach this speed. flag question: question 11
The proton accelerates from rest in a uniform electric field of 600 n/c. In order to find the time interval it takes for the proton to reach a speed of 1.50 mm/s.
We need to use the equation v = v₀ + at, where v is the final velocity, v₀ is the initial velocity (which is 0 in this case), a is the acceleration, and t is the time interval. The acceleration of the proton in the electric field is a = E/m, where E is the electric field and m is the mass of the proton. Substituting these values into the equation gives us:
1.50 mm/s = 0 + (600 n/c/1.67 x 10⁻²⁷ kg) x t
Rearranging the equation and solving for t gives us the time interval:
t = 1.50 mm/s/(600 n/c/1.67 x 10⁻²⁷ kg)
t = 8.33 x 10⁻¹³ s
t = 8.33 ms
Therefore, it takes the proton 8.33 ms to accelerate from rest to a speed of 1.50 mm/s in the uniform electric field of 600 n/c.
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at a point on the free surface of a stressed body, the normal stresses are 10 ksi (t) on a vertical plane and 31 ksi (c) on a horizontal plane. an unknown negative shear stress exists on the vertical plane. the absolute maximum shear stress at the point has a magnitude of 24 ksi. determine the principal stresses and the shear stress on the vertical plane at the point. determine the shear stress on the vertical plane. since it is stated in the problem statement that this shear stress is negative, enter a negative value.
The value of shear stress on the vertical plane is -12.25 ksi.
The given normal stress values are as follows:10 ksi (t) on a vertical plane 31 ksi (c) on a horizontal plane.Let σv and σh be the principal stresses respectively. The given unknown negative shear stress on the vertical plane is τv. The maximum shear stress value is 24 ksi. Now, let's determine the values of σv and σh using the equations,σv + σh = 10 + 31 = 41(1)σv - σh = 24∴σv = (24+41)/2 = 32.5 ksi, σh = (41-24)/2 = 8.5 ksi. Now, let's determine the shear stress on the vertical plane. The expression for maximum shear stress is given as,τmax = (σv - σh)/2 = (32.5 - 8.5)/2 = 12.25 ksi. Thus, the value of shear stress on the vertical plane is -12.25 ksi.
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The moment of inertia of a solid cylinder about its axis is given by 1/2MR 2 . If this cylinder rolls without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is:A. 1:1
B. 2:2
C. 1:2
D. 1:3
Answer:
I = 1/2 M R^2 moment of inertia
Translational energy due to rotation
Er = 1/2 I ω^2 = 1/2 M R^2 ω^2 = 1/2 M V^2 since V = R ω
Thus (A) the translational KE is equal to the rotational energy and
Ek = Er + Et for the total energy of the cylinder
-Given a capacitance of 50 nF, what resistance should your circuit have in order to have a time constant of 100 microseconds?
-From the circuit above, if you charged it to 5 Volts, then allow the circuit to discharge how long does it take to reach 1 V?
the resistance required for the circuit is 5kΩ. it takes about 2.2 microseconds for the circuit to discharge from 5 V to 1 V.
Given a capacitance of 50 nF,
the resistance that the circuit should have to have a time constant of 100 microseconds is 5kΩ.
The time constant of an RC circuit is the product of the resistance and capacitance in the circuit, according to the relationship
τ = RC.
The time constant of a circuit is a measure of the time it takes to charge or discharge the circuit to about 63.2% of its final value.
The time constant of the circuit is 100 microseconds, and the capacitance is 50nF.
Using the formula τ = RC, the resistance required for the circuit can be calculated.
To obtain the resistance required for the circuit, rearrange the formula as follows: R = τ/C
where R is the resistance, τ is the time constant, and C is the capacitance.
From the circuit above, if it is charged to 5 Volts, it takes about 2.2 microseconds to discharge to 1 V.
The time it takes for a circuit to discharge from a charged state is given by the formula:
V = V0 e^-t/RC
Where V is the voltage at any point in time,
V0 is the voltage at the start of discharge,
t is the elapsed time,
R is the resistance, and
C is the capacitance.
If the voltage is dropped to 1 V from 5 V, the voltage ratio is 1/5.
The formula for the voltage ratio is V/V0 = e^-t/RC.
Rearrange the formula as follows:-
ln(V/V0) = t/RC
When V = 1 V, V0 = 5 V, R = 5kΩ, and C = 50 nF,
substitute the values into the formula above and
solve for t.
t = -ln(1/5) RC= -ln(0.2) × 5kΩ × 50nF≈ 2.2 microseconds.
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Calculate the kinetic energy of a bullet of mass 0.015 kg, traveling at a speed of 240 m/s.
Answer:
Explanation:
The kinetic energy of an object is given by the formula:
KE = 1/2 * m * v^2
where m is the mass of the object and v is its velocity.
Plugging in the given values, we get:
KE = 1/2 * 0.015 kg * (240 m/s)^2
= 1/2 * 0.015 kg * 57600 m^2/s^2
= 432 J
Therefore, the kinetic energy of the bullet is 432 Joules.
Answer:
The kinetic energy (KE) of an object is given by the formula:
KE = (1/2) * m * v^2
where m is the mass of the object and v is its velocity.
Substituting the given values, we get:
KE = (1/2) * 0.015 kg * (240 m/s)^2
Simplifying, we have:
KE = 0.5 * 0.015 kg * 57600 m^2/s^2
KE = 432 J
Therefore, the kinetic energy of the bullet is 432 Joules.
Explanation:
The capacity of a battery to deliver charge, and thus power, decreases with temperature. The same is not true of capacitors. For sure starts in cold weather, a truck has a 500 F capacitor alongside a battery. The capacitor is charged to the full 13.8 V of the truck's battery. How much energy does the capacitor store? What is the ratio between the energy density per unit mass of the 9.0 kg capacitor system and the 130,000 J/kg of the truck's battery.
The energy stored in the capacitor is calculated as 630150 J. The ratio between the energy density per unit mass of the 9.0 kg capacitor system and the 130,000 J/kg of the truck's battery is 70.17
The formula to calculate the energy stored in a capacitor is expressed by the formula:
E = (1/2)CV²
where E is energy, C is capacitance, and V is voltage.
The question mentions that the capacitor is fully charged to 13.8 V. Therefore, the energy stored in the capacitor is given by the formula:
[tex]E = (1/2)CV^2 \\= (1/2)\times (500 F)\times {(13.8 V)}^2\\= 630150 J[/tex]
The ratio between the energy density per unit mass of the 9.0 kg capacitor system and the 130,000 J/kg of the truck's battery can be computed by dividing the energy density of the capacitor system by the energy density of the truck's battery.
We know that energy density = energy / mass of the system.
Thus, the formula to calculate the ratio is:
[tex]Ratio = \dfrac{energy density per unit mass of capacitor system}{ energy density per unit mass of truck's battery}\\Ratio= \dfrac{630150 J / 9 kg}{ 130,000 J / 1 kg}= 70.017[/tex]
Therefore, the ratio of energy density per unit mass of the capacitor system to that of the truck's battery is 70.017.
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In monsters, the allele for having one eye is dominant (A). The allele for two eyes is recessive (a ). The pedigree shows the occurrence of one eye and two eyes in four generations of a family. Label the generations and individuals 5
I can give you some broad recommendations on how to identify generations and people in a pedigree, though.
We commonly use Roman numerals (I, II, III, etc.) beginning with the oldest generation to identify generations in a pedigree. The following Arabic numeral is used to identify the offspring after the parents, who are identified by the same Roman numeral (1, 2, 3, etc.). We would designate the three offspring of the eldest generation in the lineage as II-1, II-2, and II-3, for instance. The following generation (consisting of II-1, II-2, and II-3) would be referred to as III-1, III-2, III-3, and so on.
We use Arabic numbers to identify people within a generation. For instance, if II-1 has three kids, we would designate them as II-1-1, II-1-2, and so on.
I hope this helps! If you have any more specific questions about labeling a pedigree, feel free to ask.
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What is the wavelength (in meters) of a .44 magnum bullet (20.0 grams) travelling at at 450.0 m 5-1? Remember that 1 Joule = 1 kg m? s? 1) 9.94 x 10-24 m 2) 1.49 x 10-29 m 3) 7.36 x 10-95 m O4) 1.36 x 104 m 5), 7.36 x 10-38 m
The wavelength of a .44 magnum bullet (20.0 grams) travelling at 450.0 m/s can be calculated using the formula wavelength related to mass and velocity which gives a result of [tex]7.36 \times 10^{-29} m[/tex]. Therefore, the correct answer is option(3).
According to de Broglie's wavelength equation,
λ = h/mv = h/p
Where,λ = wavelength, h = Planck's constant, m = mass of the object, v = velocity of the object, p = momentum of the object.
Given that, Mass of the bullet, m = 20.0 g = 0.020 kg
Velocity of the bullet, v = 450.0 m/s
Momentum of the bullet, p = mv = 0.020 kg × 450.0 m/s = 9.00 kg m/s
Now, using the equation for wavelength we can find:
[tex]\lambda = h/p \\= \dfrac{6.626 \times 10^{-34} J s}{ 9.00 kg m/s} \\= 7.362 \times 10^{-35} m[/tex]
Therefore, the wavelength (in meters) of a .44 magnum bullet (20.0 grams) traveling at 450.0 m/s is [tex]7.36 \times 10^{-35} m[/tex]
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max (15 kg) and maya (12 kg) are ice-skating on a frozen pond. when max is standing on the shore, he throws a 1.5-kg snowball at maya, who is standing at the center of the pond. maya catches the snowball and she and the snowball move away from the shore at 2.0 m/s. how fast was the snowball moving right before maya caught it?
The speed of the snowball before Maya caught it was 104 m/s.
According to the law of conservation of momentum, the sum of the initial momenta will be equal to the sum of the final momenta.
Mass of Max = 15 kg
Mass of Maya = 12 kg
Mass of Snowball = 1.5 kg
Now, using the law of conservation of momentum, we have
The momentum of Max + Momentum of Snowball = Momentum of Maya + Momentum of Snowball
Initial Momentum of Max = 0 (as Max is standing on the shore)
The momentum of Snowball = mv (where m is the mass of the snowball and v is the velocity of the snowball)
The momentum of Maya = mv (where m is the mass of Maya and v is the velocity of Maya with snowball)
Final Momentum of Snowball = (m + m) × v
Now putting these values, Initial momentum = 0 + 1.5 × vi = 1.5vi
Final momentum = 15 × u + 12 (2 u) = 39u (where u is the velocity of Maya with snowball after catching)
Initial Momentum = Final Momentum 1.5vi = 39u
We can write u = 2m/s
Thus putting the value of u, we can calculate the initial velocity of the snowball.
vi = u × (39 / 1.5) = 104 m/s
Thus, the speed of the snowball before Maya caught it was 104 m/s.
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A motorcyclist starts from rest and reaches a speed of 6m/s after travelling with constant acceleration for 3s. What is his acceleration?
As given, the motorcyclist starts from rest and reaches a speed of 6 m/s
after traveling with uniform acceleration for 3 seconds.
Here, initial velocity u=0
Final velocity v=6 m/s
Time t=3 sec.
Let the acceleration of the motorcycle be a.
On using the equation of motion, v=u+at
6=0+3×a
Or 3a=6
Or a=63
Or a=2 m/s2
→Therefore, the acceleration in a motorcycle is 2 m/s2.←
in a hydraulic jump occurring in a rectangular horizontal channel, the discharge per unit width is 1.5 m3/sec/m and the depth before the jump is 0.3m. estimate (a) the sequent depth (b) froude number before and after the jump. (c) energy loss (d) would the energy loss increase or decrease (and by how much) if the initial depth were changed to 0.25m?
The sequence depth is 0.36 m, froude number before and after the jump are 1.67 and 0.21. Energy loss is 0.0253 m²/s², and decrease in the energy loss is 0.0047m²/s².
What is the sequence depth?
The sequent depth (h2) of a hydraulic jump occurring in a rectangular horizontal channel can be calculated using the following formula:
h2 = (1.5/2²)/(g(h1-h2))
where, h1 = initial depth (0.3m), g = acceleration due to gravity (9.8 m/s²)
Using the formula, h2 = 0.36 m
Froude number before and after the jump:
The Froude number (Fr) is the ratio of the inertia force to the gravitational force, which can be calculated using the following formula:
Fr = (v²)/(gh²)
where, v2 = velocity after the jump (1.5m/s), h2 = sequent depth (0.36m), g = acceleration due to gravity (9.8m/s²)
Using the formula, Fr = 1.67 before the jump and 0.21 after the jump.
Energy loss: The energy loss in a hydraulic jump can be calculated using the following formula:
EL = h1g(h1-h2)b
where, h1 = initial depth (0.3m), h2 = sequent depth (0.36m), b = width of the channel (1m), g = acceleration due to gravity (9.8m/s²)
Using the formula, EL = 0.0253 m²/s²
Change in energy loss: If the initial depth (h1) is changed to 0.25m, the energy loss (EL) can be calculated using the same formula as above.
Using the formula, EL = 0.0206 m²/s²
This is a decrease in energy loss of 0.0047 m²/s².
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Place the main-sequence lifetime of each of the following stars in order from shortest to longest. (Drag and drop into the appropriate area)- Sirius A: mass 2 M_Sun, luminosity 251_Sun- Aenernar:mass 7 M_Sun, luminosity 3,150 L_Sun - The Sun: mass 1 M_Sun, luminosity 1 l_Sun- Rigel: mass 24 M_Sun, luminosity 85,000 L_Sun- Canopus: mass 8.5 M_Sun, luminosity 13,600 L_Sun- Capella A: mass 3 M_Sun, luminosity 76 L_Sun
The main sequence lifestyle of the these stars from the shortest to longest are:
Rigel: mass 24 M_Sun, luminosity 85,000Canopus: mass 8.5 M_Sun, luminosity 13,600 Achnernar: mass 7 M_Sun, luminosity 3,150Capella A: mass 3 M_Sun, luminosity 76Sirius A: mass 2 M_Sun, luminosity 251The Sun: mass 1 M_Sun, luminosity 1 How to know stars with their lifestylesStars with higher masses burn through their fuel more quickly, resulting in shorter main-sequence lifetimes.
Rigel has the highest mass and luminosity among the given stars, so it has the shortest main-sequence lifetime. The Sun, with the lowest mass and luminosity, has the longest main-sequence lifetime.
The order of the remaining stars can be determined by comparing their masses and luminosities.
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(a) When the mass is removed, the length of the cable is found to be l0=4.76m. After the mass is added, the length is measured and found to be l1=5.49m. Determine Young's Modulus Y in N/m2 for the steel cable if the weight has a mass m=35kg and the cable has a radius r=0.015m.
b) If this cable is pulled down a distance d in m from its equilibrium position it acts like a spring when released. Write an expression determining the spring constant k of this material using the cable-specific variables Y,l0,l1, and r.
To find Young's modulus Y, use [tex]Y = mg( l1 - l0 ) / ( πr^2l0 )[/tex] with given values. For the spring constant k, use [tex]k = Yπr^2 / l0, with Y, r,[/tex] and l0 given. (a) Young's modulus Y is a measure .
the stiffness of a material and is calculated using the formula Y = (mg( l1 - l0 )) / ( πr^2l0 ), where g is the acceleration due to gravity. Substituting the given values,[tex]Y = 2.08 × 10^11 N/m^2.[/tex] This means that the steel cable is relatively stiff and can resist deformation under stress. n(b) The spring constant k of the steel cable indicates its stiffness as a spring, with a higher value indicating a stiffer material that will resist deformation more strongly. In this case, the steel cable has a relatively high spring constant of 9.16 × 10^4 N/m, meaning that it will not stretch much when a force is applied.
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Use differentials to estimate the amount of material in a closed cylindrical can that is 10 cm high and 15 cm in diameter if the metal in the top and bottom is 0.1 cm thick, and the metal in the sides is 0.05 cm thick. Note, you are approximating the volume of metal which makes up the can (i.e. melt the can into a blob and measure its volume), not the volume it encloses.
The can's metal composition measured in volume is -401.94 cm^3
To estimate the amount of material in a cylindrical can, we can use differentials. Let's start by finding the volume of the can. The formula for the volume of a cylinder is:
V = πr^2h
where r is the radius of the cylinder, h is the height, and π is a constant.
The diameter of the can is 15 cm, so the radius is 7.5 cm. The height of the can is 10 cm.
First, we need to find the volume of the metal in the top and bottom of the can. The thickness of the metal is 0.1 cm, so the radius of the top and bottom of the can is reduced by 0.1 cm. Therefore, the volume of the metal in the top and bottom is:
V_top&bottom = π(7.4)^2(0.1) ≈ 16.31 cm³
Next, we need to find the volume of the metal in the sides of the can. The thickness of the metal is 0.05 cm, so the radius of the sides of the can is reduced by 0.1 cm. Therefore, the volume of the metal in the sides of the can is:
V_sides = π(7.4)^2(10) ≈ 2153.78 cm³
The total volume of the can is:
V_total = π(7.5)^2(10) ≈ 1767.15 cm³
To find the volume of the metal that makes up the can, we subtract the volume of the empty space inside the can from the total volume of the can:
V_metal = V_total - V_empty
V_empty = V_top&bottom + V_sides ≈ 2169.09 cm³
Therefore, the volume of the metal that makes up the can is:
V_metal ≈ 1767.15 cm³ - 2169.09 cm³ ≈ -401.94 cm³
Since this result is negative, it does not make sense in the context of the problem. This suggests that there may be an error in our calculations, possibly due to the approximations made when using differentials. Nevertheless, we can use this method to estimate the amount of material in the can, although we may need to use more accurate methods for precise measurements.
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calculate the electrostatic force between 1nc and 1nc charges at a distance of 1 m from each other. do not forget to mention the direction of the force, too.
The electrostatic force between 1nc and 1nc charges at a distance of 1 m from each other is 9.0 × 10^-9 N. The direction of the force is given by Coulomb's law and is along the line joining the two charges. It is either repulsive or attractive based on the type of the charges.
What is Coulomb's law? Coulomb's law is an equation used to calculate the electrostatic force between two charged particles. According to this law, the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. The equation for Coulomb's law is given by:
F = k * (q1 * q2) / r^2
Where F is the electrostatic force,k is Coulomb's constant,q1 and q2 are the charges of the particles, and r is the distance between the particles.
Given,
Charge of particle 1, q1 = 1 nc
Charge of particle 2, q2 = 1
distance between particles, r = 1
coulomb's constant, k = 9 × 10^9 N m^2/C^2
Now, we can use Coulomb's law to calculate the electrostatic force between the two charges. Substituting the given values in the equation:
F = k * (q1 * q2) / r^2= 9 × 10^9 N m^2/C^2 * (1 × 10^-9 C) * (1 × 10^-9 C) / (1 m)^2= 9.0 × 10^-9 N
Thus, the electrostatic force between 1nc and 1nc charges at a distance of 1 m from each other is 9.0 × 10^-9 N. The direction of the force is given by Coulomb's law and is along the line joining the two charges. It is either repulsive or attractive based on the type of the charges.
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a 38.6 lb weight is supported on several springs whose combined stiffness is 6.4 lb/in. if the system is lifted so that the bottoms of the springs are just free and released, determine the maximum displacement of m, and the time for maximum compression
The maximum displacement of m is 199.14, and the time for maximum compression 1.56 seconds.
Given:
Weight, W = 38.6 lb
K(combined stiffness) = 6.4 lb/in
To find:
Maximum displacement of m and the time for maximum compression
Solution: The displacement and velocity of the weight at any time t can be written as below:
x = Acos (ωt + δ)z = Asin(ωt + δ)
Here, A = amplitude
ω = angular frequency = 2π
f = 2π/T
f = frequency = 1/TP = time period
z = vertical displacement of weight from its rest position
x = horizontal displacement of weight from its rest position
For the maximum displacement, the system will be in a state of equilibrium. i.e. ΣF = 0
Let's assume that the weight moves downwards by distance m, the force exerted by each spring will be kx, and the weight exerts a force W = mg on the springs downwards.
Here, m = 38.6 lbs, g = 32.2 ft/s2 and k = K/m = 6.4/38.6 = 0.1657 lb/in
ΣF = -kx - kx - kx - kx - kx - kx + mg = 0-6.4m = -38.6 * 32.2m = 199.14 in (Maximum Displacement of M)The maximum compression will occur when the weight is at the lowest point, i.e. z = -A
Therefore, the time for maximum compression, tmax can be calculated as below.
z = Asin(ωt + δ)At the point of maximum compression, t = tmax
z = -A = -199.14 in (as calculated above)
Therefore,-199.14 = Asin(ωtmax + δ)
Here, A = kx = 6.4×199.14/32.2 = 39.45 inω = 2π/T = 2πf = 2π/4.72 = 1.33 rad/s (where T = time period and f = frequency)
Therefore,-199.14 = 39.45sin(1.33tmax + δ)sin(1.33tmax + δ) = -5.05tmax = 1.56 s
Thus, the maximum displacement of m is 199.14 inches and the time for maximum compression is 1.56 seconds.
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a 1-kg ball is thrown at 10 m/s straight upward. neglecting air resistance, the net force that acts on the stone when it is halfway to the top of its path is about
The net force that acts on the stone when it is halfway to the top of the path is zero, this is because of the gravitational acceleration.
Why does the net force turn zero?When a 1 kg ball is thrown at 10 m/s straight upward, and air resistance is ignored, the net force that acts on the stone when it is halfway to the top of its path is approximately zero.
Since air resistance is ignored, the net force acting on the 1-kg ball is just the force due to gravity. In the absence of air resistance, the acceleration of the ball will remain constant and equal to g, which is -9.81 m/s², because the weight of the ball is mg, where m is the mass of the ball and g is the gravitational acceleration.
So, the net force acting on the ball will be given by:
Net force = m × g = (1 kg) × (-9.81 m/s²) = -9.81 N
Therefore, the net force acting on the ball when it is halfway to the top of its path is zero.
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Why are masses listed on the periodic table not whole #'s. Ex. 15.9999 for oxygen?
The masses listed on the periodic table are not whole numbers because they represent the weighted average of all the naturally occurring isotopes of an element.
What are Isotopes ?Isotopes are atoms of the same element that have different numbers of neutrons in their nuclei, resulting in slightly different masses. Since the abundance of each isotope in nature can vary, the weighted average takes into account the abundance of each isotope and their corresponding masses, resulting in a decimal value. For example, oxygen has three naturally occurring isotopes, with mass numbers of 16, 17, and 18.
Why only O-16 isotopes ?The most abundant isotope is oxygen-16, but the other isotopes are also present in trace amounts, leading to a weighted average of 15.9994 amu (atomic mass units). This is why the mass listed on the periodic table for oxygen is 15.999, which is a rounded value of the weighted average.
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The masses listed on the periodic table are not whole numbers because they represent the average atomic mass of all the naturally occurring isotopes of an element, taking into account their relative abundances.
What are isotopes ?
Isotopes are atoms of the same element that have different numbers of neutrons in their nucleus, which affects their atomic mass. Some isotopes of an element are more abundant than others, and their relative abundances are taken into account when calculating the average atomic mass.
For example, oxygen has three naturally occurring isotopes: oxygen-16, oxygen-17, and oxygen-18. Oxygen-16 is the most abundant isotope, making up about 99% of all oxygen atoms. Oxygen-17 and oxygen-18 are much less abundant, but they still contribute to the overall atomic mass of the element.
The atomic mass listed on the periodic table for oxygen (15.9994) is the weighted average of the atomic masses of all three isotopes, taking into account their relative abundances. This average is not a whole number because the isotopes have different atomic masses and abundances, and their contributions to the overall average are weighted accordingly.
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how would we get mercury to be reclassified as a minor body?
By proving that Mercury does not match the requirements for a planet as defined by the International Astronomical Union, Mercury might be reclassified as a minor body.
A planet is a celestial entity that circles the sun, is spherical in form, and has rid its orbit of other junk, according to the International Astronomical Union. Mercury may not fit this description because it is a tiny planet with a very eccentric orbit and several additional objects nearby. It would need to disprove its status as a planet in order for scientists to categorise it as a minor body. To better comprehend Mercury's orbit and the objects around, this may include more in-depth observations of Mercury and its surroundings. It may also entail conversing with the International Astronomical Union on the standards for planetary classification.
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A projectile is fired from ground level with a speed of 150 m/s at an angle 30° above the horizontal on an airless planet where g = 10.0 m/s2. What is the horizontal component of its velocity after 4.0 s?
The horizontal component of its velocity after 4.0 s is 129.90 m/s after the projectile is fired from ground level with a speed of 150 m/s at an angle 30° above the horizontal on an airless planet where g = 10.0 m/s2.
What is the horizontal component of its velocity after 4.0 s? To find the horizontal component of its velocity after 4.0 seconds, we have to first find the initial horizontal velocity of the projectile as it was fired at an angle of 30° above the horizontal. We can use trigonometric ratios for that.
Hence, the initial horizontal component of the velocity = Vcosθ=150 cos 30°=150 × √3/2=129.90 m/s.
The vertical component of velocity is given by: Vsinθ=150 sin 30°=75.0 m/s.
Now, we can use the formula for the horizontal displacement of a projectile to find its horizontal velocity after 4 seconds, horizontal displacement of projectile= Vcosθ × t
So, the horizontal displacement of the projectile after 4 seconds= 129.90 × 4= 519.6 m
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Need help on some of my homework please!
Answer:
Carbon - B (Atomic mass of 12)
Oxygen- C (has eight protons)
silicon - A (atomic mass of 28)
Sulfur - D (atomic number of 16)