Answer:
the magnitude of the average force on the bumper is 3189.8 N
Explanation:
Given the data in the question;
In terms of force and displacement, work done is;
W =[tex]F^>[/tex] × [tex]x^>[/tex]
W = [tex]Fxcos\theta[/tex] ------- let this be equation 1
where F is force applied, x is displacement and θ is angle between force and displacement.
Now, since the displacement of the bumper and force acting on it is in the same direction,
hence, θ = 0°
we substitute into equation 1
W = [tex]Fxcos([/tex] 0° [tex])[/tex]
W = [tex]Fx[/tex] ------- let this be equation 2
Now, using work energy theorem,
total work done on the system is equal to the change in kinetic energy of the system.
[tex]W_{net[/tex] = ΔKE
= [tex]\frac{1}{2}[/tex]mv² - [tex]\frac{1}{2}[/tex]mu² --------- let this be equation 3
where m is mass of object, v is final velocity, u is initial velocity.
from equation 2 and 3
[tex]Fx[/tex] = [tex]\frac{1}{2}[/tex]mv² - [tex]\frac{1}{2}[/tex]mu²
we make F, the subject of formula
F = [tex]\frac{m}{2x}[/tex]( v² - u² )
given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s
so we substitute
F = [tex]\frac{830}{(2)(0.255)}[/tex]( (0)² - (1.4)² )
F = 1627.45098 ( 0 - 1.96 )
F = 1627.45098 ( - 1.96 )
F = -3189.8 N
The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.
Therefore, the magnitude of the average force on the bumper is 3189.8 N
a camera employs _lens to form_images
Answer:
a camera employs camera lens to firm some images.
Explanation:
hope this helps.
.Use Newton's third law to describe the forces that are exerted by the falling egg and the ground. Explain how the use of the straws in the design affects the forces
Answer:
Newton's third law states that for every action, there is an equal and opposite reaction. That means when you exert a force on an object, the object exerts a force back on you. ... Using shock-absorbing materials can help reduce the amount of force exchanged between the ground and the egg.
Explanation:
I hope it helps you
When one of the tall straws hits the ground, the energy is transferred to the center of the pyramid and then to the egg, but as the middle straw is connected to the outer surface of the egg, energy enters trying to make the egg rotate. This is governed by Newton's third law.
What is Newton's third law?If an object exerts a force on another object, then another object must exert a force of equal magnitude and opposite direction back on first object.
What are examples of Newton's third law?Examples of Newton's third law are:
A swimmer moves forward by pushing off the side of pool. This way, the wall pushes in opposite direction and giving acceleration.
Another example is rockets move forward by expelling gas backward at high velocity. This means the rocket exerts a large backward force on the gas in the rocket combustion chamber, and the gas therefore exerts a large reaction force forward on the rocket. This reaction force is called thrust.
Helicopters similarly create lift by pushing air down, thereby experiencing an upward reaction force. Birds and airplanes also fly by exerting force on air in a direction opposite to that of whatever force they need. For example, the wings of a bird force air downward and backward in order to get lift and forward motion.
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a 50kg skater on level ice, has built up her speed to 30km/h. how far will she coast before sliding friction dissipates her energy?
Answer:
belpw
Explanation:
The distance prior to the sliding friction dispersing her energy would be:
- The distance will remain unaffected by the sliding friction i.e. 354m
As we know, When Sliding friction dissolves her energy, leading her Kinetic Energy to turn 0 on coming to the state of rest. So,
[tex]1/2 mv^2 - 1/2 mu^2 = -W[/tex] (∵ Work in -ve denotes it is done opposite to friction)
Given that,
m(mass) [tex]= 50 kg[/tex]
v(velocity) [tex]= 30 km/hr[/tex] or [tex]8.33 m/s[/tex]
The coefficient of Kinetic Friction [tex]= 0.01[/tex]
g(gravitational force) [tex]= 9.8 m/s^2[/tex]
Initial Velocity(u) [tex]= 30[/tex] × [tex]1000/3600 m/s[/tex]
[tex]= 8.33 m/s[/tex]
Now by employing the provided values,
[tex]F =[/tex] μ[tex]mg[/tex]
[tex]= (0.01) (50) (9.8)[/tex]
[tex]= 4.9[/tex]
∵ [tex]F = 4.9 N[/tex]
By using the above expression, we will find the distance;
[tex]1/2 mv^2 - 1/2 mu^2 = -W[/tex]
⇒ [tex]1/2 (50) (0)^2 - 1/2 (50) (8.33)^2 = -4.9(S)[/tex]
⇒ [tex]1734.7225 = 4.9S[/tex]
⇒ [tex]S = 1734.7225/4.9[/tex]
∵ [tex]S = 354 m[/tex]
Because [tex]1/2 mv^2 - 1/2 mu^2 = -W[/tex] [tex]= -[/tex] μmgS
⇒ [tex]S = (u^2 - v^2)[/tex]/2μ[tex]g[/tex]
Thus, the distance will remain unaffected by the sliding friction i.e. 354m
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A 5.0-kg solid cylinder of radius 0.25 mis free to rotate about an axle that runs along the cylinders length and passes through its center. A thread wrapped around the cylinder is weighed down by a mass of 2.0 kg so as to unwrap and make the cylinder rotate as this mass falls. Ignore any friction in the axle. If there is no slippage between the thread and the cylinder, and the cylinder starts from rest (a) Calculate the velocity of the block after it has fallen a distance of 2.0m. Give your answer in m.s (b) Calculate the total work done by the rope on the cylinder after the block has fallen a distance of 2.0 m. Give your answer in Joule.
Answer:
157n is the correct answer
a ball is projected horizontally from the top of a hill with a velocity of 30 m per second if it reaches the ground 5 seconds later the height of the hill is
Answer:
The height of the hill is 125m.
Explanation:
Since the ball is projected horizontally from the top of a hill, there is no vertical component of the velocity of projection. Therefore, so far as motion in the vertical direction is concerned, the ball is just dropped, ie its initial velocity u is merely 0 m/s. It reaches the ground after 5 seconds from the moment of projection, under the action of accerelation due to gravity.
Using,
s = u t + ½ a t². In this expression u= 0 m/s, a = 10m/s², t = 5 s. Substituting in the equation we get,
s = 0× t + ½ ×10 m/s²× 5²s²= 5× 25 m = 125m.
A ceramic tile measuring 50 cm x50cm has been designed to bear a pressure of 40 N/in . Will it with stand a force of 5 N?
Answer:
The ceramic tile will stand the force of 5 N.
Explanation:
Step 1: Calculate the area (A) of the ceramic tile
We will use the formula for the area of a square.
A = 50 cm × 50 cm = 2500 cm²
Step 2: Convert "A" to in²
We will use the conversion factor 1 in² = 6.45 cm².
2500 cm² × 1 in²/6.45 cm² = 3.9 × 10² in²
Step 3: Calculate the pressure (P) exerted by a force (F) of 5 N
We will use the following expression.
P = F/A
P = 5 N / 3.9 × 10² in² = 0.013 N/in²
Since the pressure exerted would be less than the maximum pressure resisted (40 N/in²), the ceramic tile will stand the force of 5 N.
An electron is moving through an (almost) empty universe at a speed of 628 km,/s toward the only other object in the universe — an insulating sphere with a diameter of 4 m and charge density 3nC/m2 on its outside surface. The sphere "captures" the electron, which falls into a circular orbit.
Required:
Find the radius and period of the orbit.
Answer:
r = 2,026 10⁹ m and T = 2.027 10⁴ s
Explanation:
For this exercise let's use Newton's second law
F = m a
where the force is electric
F = [tex]k \frac{q_1q_2}{r^2}[/tex]
Acceleration is centripetal
a = v² / r
we substitute
[tex]k \frac{q_1q_2}{r^2} = m \frac{v^2}{r}[/tex]
r = [tex]k \frac{q_1q_2}{m \ v^2}[/tex] (1)
let's look for the charge in the insulating sphere
ρ = q₂ / V
q₂ = ρ V
the volume of the sphere is
v = 4/3 π r³
we substitute
q₂ = ρ [tex]\frac{4}{3}[/tex] π r³
q₂ = 3 10⁻⁹ [tex]\frac{4}{3}[/tex] π 4³
q₂ = 8.04 10⁻⁷ C
let's calculate the radius with equation 1
r = 9 10⁹ 1.6 10⁻¹⁹ 8.04 10⁻⁷ /(9.1 10⁻³¹ 628 10³)
r = 2,026 10⁹ m
this is the radius of the electron orbit around the charged sphere.
Since the orbit is circulating, the speed (speed modulus) is constant, we can use the uniform motion ratio
v = x / t
the distance traveled in a circle is
x = 2π r
In this case, time is the period
v = 2π r /T
T = 2π r /v
let's calculate
T = 2π 2,026 10⁹/628 103
T = 2.027 10⁴ s
A certain electric stove has a 16 Ω heating element. The current going through the element is 15 A. Calculate the voltage across the element.
The voltage across the element is = 240 V
I hope you understand....
Mark me as brainliest....
Thanks...
If we warm a volume of air, it expands. Does it then follow that if we expand a volume of air, it warms? Explain.
Answer:
nope don't think so
Explanation:
the heat causes the molecules to move faster therefore expanding in watever it the air is in
Answer:
If the pressure is maintained at a constant value then both statements are equivalent:
P V = n R T Ideal gas equation
if P, n, and R are maintained at constant values then
V = T and the two expressions are equivalent
You are a venture capitalist that is asked to invest in a startup company that claims it will be able to launch tiny "micro space probes" into space at close to the speed of light using a massive electromagnetic rail gun system2. You are cynical about their cost estimates and decide to analyze the problem in more detail before you invest in their company. Neglect air resistance for this worksheet.
1. A typical payload they claim to launch will weigh 1 kilogram and be accelerated to 90% the speed of light. How much electrical energy will the rail gun require to launch the probe, assuming it is 20% efficient at converting electrical energy to projectile kinetic energy?
2. Given that typical electrical costs are about 5 cents/MJ, how much would this launch cost? Would you invest in this company?
43. You are also concerned about safety. What happens if this projectile were to hit an airplane that is flying overhead and dissipate all of its kinetic energy in the collision? To give you a sense of scale, a large nuclear explosion generates about 1015 J of energy.
4. The system must be able to launch probes to all parts in the sky and must be transportable on a ship. Assume that the railgun is mounted on a frigate-class navy ship (weight = 4,000 metric tons).
a. Will the recoil momentum of the ship be relativistic? Justify your argument.
b. At what speed will the ship recoil after it launches a probe? Do you think that this is a problem for the ship?
Answer:
1. 5.825 × 10¹⁷ J
2. i. $ 29.125 billion ii. I would not invest in the company
3. A nuclear reaction would occur if the payload hits an airplane flying overhead and dissipates all its kinetic energy in the collision.
4. a i. The momentum will not be relativistic
ii. This is because objects with large masses do not move at relativistic speeds
b i. 155 m/s
ii. This speed wouldn't be a problem for the ship.
Explanation:
1. A typical payload they claim to launch will weigh 1 kilogram and be accelerated to 90% the speed of light. How much electrical energy will the rail gun require to launch the probe, assuming it is 20% efficient at converting electrical energy to projectile kinetic energy?
The kinetic energy of the payload is K = (γ - 1)mc² where m = mass of payload = 1 kg, c = speed of light = 3 × 10⁸ m/s and γ = 1/√(1 - β²) where β = 0.9 (since the payload moves at 90 % speed of light)
So, K = (γ - 1)mc²
= (1/√(1 - β²) - 1)mc²
= (1/√(1 - (0.9)²) - 1) × 1 kg × (3 × 10⁸ m/s)²
= (1/√(1 - 0.81) - 1) × 1 kg × (3 × 10⁸ m/s)²
= (1/√0.19 - 1) × 1 kg × (3 × 10⁸ m/s)²
= (1/0.436 - 1) × 1 kg × (3 × 10⁸ m/s)²
= (2.294 - 1) × 1 kg × (3 × 10⁸ m/s)²
= 1.294 × 1 kg × 9 × 10¹⁶ m²/s²
= 11.65 × 10¹⁶ kgm²/s²
= 1.165 × 10¹⁷ J
Let E be the total electrical energy of the rail gun. Since 20 % of this energy is converted to kinetic energy of the payload, we have
20 % of E = K
0.2E = K
E = K/0.2
= 1.165 × 10¹⁷ J/0.2
= 5.825 × 10¹⁷ J
2 Given that typical electrical costs are about 5 cents/MJ, how much would this launch cost? Would you invest in this company?
i. how much would this launch cost?
Since the total energy required is E = 5.825 × 10¹⁷ J = 5.825 × 10¹¹ MJ and it costs 5 cent/MJ. So the total cost of energy will be total energy rate = 5.825 × 10¹¹ MJ × 5 cent/MJ = 29.125 × 10¹¹ = 2.9125 × 10¹² cents. Converting this to dollars, we have 2.9125 × 10¹² cents/100 cents/dollar = 2.9125 × 10¹⁰ dollars = 29.125 × 10⁹ dollars = 29.125 billion dollars = $ 29.125 billion
ii. Would you invest in this company?
I would not invest in the company
3. You are also concerned about safety. What happens if this projectile were to hit an airplane that is flying overhead and dissipate all of its kinetic energy in the collision? To give you a sense of scale, a large nuclear explosion generates about 1015 J of energy.
Since the kinetic energy of the payload is 1.165 × 10¹⁷ J and a nuclear explosion generates about 10¹⁵ J of energy, then a nuclear reaction would occur if the payload hits an airplane flying overhead and dissipates all its kinetic energy in the collision.
4. The system must be able to launch probes to all parts in the sky and must be transportable on a ship. Assume that the rail gun is mounted on a frigate-class navy ship (weight = 4,000 metric tons).
a. Will the recoil momentum of the ship be relativistic? Justify your argument.
i. Will the recoil momentum of the ship be relativistic?
The momentum will not be relativistic.
ii. Justify your argument.
This is because objects with large masses do not move at relativistic speeds. Since the speed cannot be relativistic, its momentum which is the product of mass and speed is non-relativistic
b. At what speed will the ship recoil after it launches a probe? Do you think that this is a problem for the ship?
i. At what speed will the ship recoil after it launches a probe?
Since the total energy of the payload E' = K + mc² = 1.165 × 10¹⁷ J + 1 kg × (3 × 10⁸ m/s)² = 1.165 × 10¹⁷ J + 1 kg × 9 × 10¹⁶ m²/s² = 11.65 × 10¹⁶ J + 9 × 10¹⁶ J = 20.65 × 10¹⁶ J
Also, E'² = (pc)² + (mc²)² where p = momentum of payload
So, making p subject of the formula, we have
(pc)² = E'² - (mc²)²
pc = √[E'² - (mc²)²]
p = √[E'² - (mc²)²]/c
substituting the values of the variables into the equation, we have
p = √[E'² - (mc²)²]/c
p = √[(20.65 × 10¹⁶ J)² - 1kg × (3 × 10⁸ m/s²)²]/3 × 10⁸ m/s
p = √[(20.65 × 10¹⁶ J)² - (1kg × 9 × 10⁸ m²/s²)²]/3 × 10⁸ m/s
p = √[426.4225 × 10³² J² - 81 × 10³² J²]/3 × 10⁸ m/s
p = √[345.4225 × 10³² J²]/3 × 10⁸ m/s
p = 18.59 × 10¹⁶/3 × 10⁸ m/s
p = 6.20 × 10⁸ kgm/s
From the law of conservation, this momentum of the payload equals the momentum of recoil of the ship.
So, p = m'v where m' = mass of navy ship = 4,000 metric tons = 4,000 × 1000 kg = 4 × 10⁶ kg and v = speed of navy ship
So, v = p/m'
= 6.20 × 10⁸ kgm/s ÷ 4 × 10⁶ kg
= 1.55 × 10² m/s
= 155 m/s
ii. Do you think that this is a problem for the ship?
Since the ship's speed is 155 m/s, which is small for an object with such a large mass, this speed wouldn't be a problem for the ship.
The phrase "hot air rises" describes which type of heat transfer?
A. Conduction
B. Insulation
C. Convection
D. Radiation
Answer:
Option C
Explanation:
C. Convection...
The phrase 'hot air rises'describe the convection as a type of heat transfer therefore the correct answer is option C.
It is the type of heat transfer in which the heat is transferred from the bulk motion of the particles of liquid and gases due to the presence of temperature gradient.
The hot air rises is one of the best example of convection currents
The boiling water is also an example of convection heat transfer process in which the hot water rises to the top of the surface due to the bulk molecular movement of convection current .
The hot air rides is the bulk movement of the gas molecules because of the presence of the temperature gradients resulting in the varied density of air ,the hot air rising to the surface have less density as compared to the air settling near the surface ,the convection current phenomenon is responsible for the phrase "hot air rises" ,therefore the correct option is the option C.
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A 10,000J battery is depleted in 2h. What power consumption is this? *
A) 5000W
B) 3W
C) 1.4W
D) 20000W
show your work please
Answer:
C) 1.4W
Explanation:
Given;
energy of the battery, E = 10,000 J
duration of the battery, t = 2 hours
The power consumption of the battery is calculated as;
[tex]Power = \frac{Energy}{time} \\\\Power = \frac{10,000}{2 \times 3600 } \\\\Power = 1.4 \ J/s = 1.4 \ W[/tex]
Therefore, the power consumption of the battery is 1.4 W
The fact that we can define electric potential energy means that:
A) the electric force is nonconservative
B) the electric force is conservative
C) the work done on a charged particle depends on the path it takes
D) there is a point where the electric potential energy is exactly zero
E) it takes work for the electric force to move from some point a to some other point b and back again
Answer: I'd go with Option B if you're allowed to pick One Option Only.
Option A is wrong. Electric Forces arent Non - Conservative. They're Conservative forces.
Option B: We Can Only Define Potential Energies for conservative Forces/fields✅
Option C is wrong. Since the Electric Force is Conservative... It doesn't depend on the path taken.
Option D... There's a Point where the electric Potential Energy is Zero but the separation Distance Would be large.
E is wrong because the Workdone around a closed path is Zero.
The electric potential energy results from the conservative Coulomb forces.
What is electric potential energy?The electric potential energy is the potential energy that results due to the conservative Coulomb forces and is always associated with the set of two charged particles.
Any object can have electric potential energy by two key elements first is its own charge and the second is its position with respect to the other charged particle.
The electric potential energy-containing only one point charge will be zero. Because there are no other sources of electrostatic force against which an external agent must do work in moving the point charge from infinity to its final position.
Thus the electric potential energy results from the conservative Coulomb forces.
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A 0.277 kg bat is flying 4.25 m/s.
It then doubles its velocity.
How much KE did it gain?
Answer:
7.51
Explanation:
The mass of the bat is given to us as:
m = 0.277 kg
,
and its initial velocity is:
v
i = 4.25 m/s
When the velocity is doubled, the new value becomes:
v f = 4.25 m/s × 2 = 8.5 m/s
.
The formula for kinetic energy is:
K
=
1
2
m
v
2
.
The change in kinetic energy is:
Δ K = 1 /2( m ) ( v2 f − v 2 i ) = 1 2 ( 0.277)
10-2.5=7.5J
7.51 J
.
The bat gained 7.51 joules.
An exoplanet has three times the mass and one-fourth the radius of the Earth. Find the acceleration due to gravity on its surface, in terms of g, the acceleration of gravity at Earth's surface. A planet's gravitational acceleration is given by gp = G Mp/r^2p
a. 12.0 g.
b. 48.0 g.
c. 6.00 g.
d. 96.0 g.
e. 24.0 g.
Answer:
b. 48.0 g.
Explanation:
Given;
mass of the exoplanet, [tex]M_p = 3M_e[/tex]
radius of the exoplanet, [tex]r_p = \frac{1}{4} r_e[/tex]
The acceleration due to gravity of the planet is calculated as;
[tex]g_p = \frac{GM_p}{r_p^2} \\\\for \ Earth's \ surface\\\\g = \frac{GM_e}{r_e^2} \\\\G = \frac{gr_e^2}{M_e} = \frac{g_pr_p^2}{M_p} \\\\\frac{gr_e^2}{M_e} = \frac{g_p(\frac{r_e}{4}) ^2}{3M_e} \\\\\frac{gr_e^2}{M_e} = \frac{g_pr_e ^2}{16\times 3M_e} \\\\g = \frac{g_p}{48} \\\\g_p = 48 \ g[/tex]
Therefore, the correct option is b. 48.0 g
Describe how the constant electric potential from the battery drives the charges in the conducting wire in your own words.
Explanation:
It is known that charges tend to move from a higher potential towards lower potential. As a result, charges tend to move from a positive terminal to negative terminal of the battery.
For example, at a certain height water is stored in a water tank. This height is correspondent to the constant potential and water contained is the charge present in the battery.
When a pipe is connected from the tank to the ground then water will start moving towards the ground and similarly if a conducting wire is connected across the battery then charge will also begin to flow in the same direction which is called current.
What is the electric field 3.9 m from the center of the terminal of a Van de Graaff with a 6.60 mC charge, noting that the field is equivalent to that of a point charge at the center of the terminal
Answer:
the electric field is 3.91 x 10⁶ N/C
Explanation:
Given the data in the question;
Electric field at a point due to point charge is;
E = kq/r²
where k is the constant, r is the distance from centre of terminal to point where electric field is, q is the excess charge placed on the centre of terminal of Van de Graff,a generator
Now, given that r = 3.9 m, k = 9.0 x 10⁹ Nm²/C², q = 6.60 mC = 6.60 x 10⁻³ C
so we substitute into the formula
E = [(9.0 x 10⁹ Nm²/C²)( 6.60 x 10⁻³ C)] / ( 3.9 )²
E = 59400000 / 15.21
E = 3.91 x 10⁶ N/C
Therefore, the electric field is 3.91 x 10⁶ N/C
Mary applies a force of 73 N to push a box with an acceleration of 0.48 m/s^2. When she increases the pushing force to 84 N, the box's acceleration changes to 0.64 m/s^2. There is a constant friction force present between the floor and the box.
Required:
a. What is the mass of the box?
b. What is the coefficient of kinetic friction between the floor and the box?
Answer: [tex]68.75\ kg, 0.06[/tex]
Explanation:
Mary applies a force of 73 N to create an acceleration of [tex]0.48\ m/s^2[/tex]
When She increases force to 84 N, it creates an acceleration of [tex]0.64\ m/s^2[/tex]
Friction opposes the motion of box
[tex]\Rightarrow 73-f=m\times 0.48\quad \ldots(i)\\\Rightarrow 84-f=m\times 0.64\quad \ldots(ii)[/tex]
Subtract (i) from (ii)
[tex]\Rightarrow 11=m(0.64-0.48)\\\Rightarrow m=68.75\ kg[/tex]
Therefore friction is
[tex]\Rightarrow f=73-68.75\times 0.48\\\Rightarrow f=73-33\\\Rightarrow f=40\ N[/tex]
Here, friction is kinetic friction which is given by
[tex]\Rightarrow f=\mu_kmg\\\Rightarrow 40=\mu_k 68.75\times 9.8\\\Rightarrow \mu_k=0.061[/tex]
The ejection seat has an acceleration of 8gees (8xgravity or ~80m/s/s). He has a mass of 70kg. The total force on him from the chair/rocket would be.
A) (80m/s/s)(70kg)=5600N
B) (80m/s/s)(70kg)=5600N + Fg = 5600N+(70kg)(9.8N/kg)~5600N+700N=6300N
C) (80m/s/s)(70kg)=5600N - Fg = 5600N+(70kg)(9.8N/kg)~5600N-700N=4900N
D) I need the time
Answer:
B) (80m/s/s)(70kg)=5600N + Fg = 5600N+(70kg)(9.8N/kg)~5600N+700N=6300N
Explanation:
The total force by the chair is given by the following formula:
[tex]F = m(g+a)[/tex]
where,
F = Force = ?
m = mass of person = 70 kg
g = value of acceleration dueto gravity = 9.81 m/s²
a = acceleration of ejection seat = 8g = 80 m/s²
Therefore,
[tex]F = mg+ma \\F = (70\ kg)(9.8\ m/s^2)+(70\ kg)(80\ m/s^2)\\F = 6300 N[/tex]
Therefore, the correct option is:
B) (80m/s/s)(70kg)=5600N + Fg = 5600N+(70kg (9.8N/kg)~5600N+700N=6300N
A permanent magnet is pushed into a wire, left there for a while, and then pulled out. During which time does a current run though the wire? A from the time that the magnet is pushed into the coil to the time it is pulled out B while the magnet remains within the coil C while the magnet is moving D only while the magnet is being pulled out of the coil
Answer:
C. while the magnet is moving
Explanation:
Electromagnetic induction implies the production of electric current by mere movement of a magnet with respect to a coil or wire.
In the given question, current would be induced in the wire only when the magnet moves. That is either when the magnet is pushed into a wire, or when pulled out. But no current would flow through the wire when the magnet is left there for a while.
The current is induced because of the motion involved. Thus, the appropriate option is C.
A charge Q is distributed uniformly around the perimeter of a ring of radius R. Determine the electric potential difference between the point at the center of the ring and a point on its axis at a distance 6R from the center. (Use any variable or symbol stated above along with the following as necessary: ke.)
ΔV = V(0) − V(6R) = ?
Answer:
the electric potential difference between the point at the center of the ring and a point on its axis ΔV is [tex]( 0.8356 )[/tex][tex]\frac{kQ}{R}[/tex]
Explanation:
Given the data in the question;
electric potential at the center of the ring V₀ = kQ / R
electric potential on the axis point Vr = kQ / √( R² + x² )
at a distance 6R from the center,
point at x = 6R
so distance circumference r = √( R² + (6R)² )
so
electric potential on the axis point Vr = kQ / √( R² + (6R)² )
Vr = kQ / R√37
Now
ΔV = V₀ - Vr
we substitute
ΔV = ( kQ / R) - ( kQ / R√37 )
ΔV = kQ/R( 1 - 1/√37 )
ΔV = kQ/R( 1 - 0.164398987 )
ΔV = kQ/R( 0.8356 )
ΔV = [tex]( 0.8356 )[/tex][tex]\frac{kQ}{R}[/tex] { where k = [tex]\frac{1}{4\pi e_0}[/tex] }
Therefore, the electric potential difference between the point at the center of the ring and a point on its axis ΔV is [tex]( 0.8356 )[/tex][tex]\frac{kQ}{R}[/tex]
human activities that interfere with distribution of natural resources are contributing to the increase of earthquake risk.
Answer:
Changes in climate conditions, especially the warming of global temperatures increases the likelihood of weather-related natural disasters. ... This is most visible when seen through changes in the intensity and frequency of droughts, storms, floods, extreme temperatures and wildfires.
Describe an experiment to find the density of copper turning using a density bottle and kerosene
The density is the ratio of mass to volume of a substance.
What is the density bottle?The density bottle is used to obtain the density of substance by measuring the volume of the fluid displaced.
If the mass of copper turnings are previously weighed and known, the volume of the fluid displaced in the density bottle is the volume of the copper turning.
Hence;
Density = mass/ volume
Learn more about density:https://brainly.com/question/15164682?
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A man pulls his dog (m=20kg) on a sled with a force of 100N at a 60° angle from the horizontal. What is the horizontal component of the force?
A) 100N
B) 196N
C) 50N
D) 86N
show your work please
Answer:
the horizontal component of the force is 50 N
Explanation:
Given;
force applied by the man, F = 100 N
angle of inclination of the force, θ = 60⁰
mass of the dog, m = 20 kg
The horizontal component of the force is calculated as;
[tex]F_x = F\times cos(\theta)\\\\F_x = 100 \ N \times cos(60^0)\\\\F_x = 100\ N \times 0.5\\\\F_x = 50 \ N[/tex]
Therefore, the horizontal component of the force is 50 N
What relationship do you see between a star colour and temperature
Answer:
Stars emit colors of many different wavelengths, but the wavelength of light where a star's emission is concentrated is related to the star's temperature - the hotter the star, the more blue it is; the cooler the star, the more red it is
Answer:
Stars emit colors of many different wavelengths, but the wavelength of light where a star's emission is concentrated is related to the star's temperature – the hotter the star, the more blue it is; the cooler the star, the more red it is.
Explanation:
Pa brainliest po ty
Part A What will be the equilibrium temperature when a 227 g block of copper at 283 °C is placed in a 155 g aluminum calorimeter cup containing 844 g of water at 14.6°C?
Answer:
T = 20.84°C
Explanation:
From the law of conservation of energy:
Heat Lost by Copper Block = Heat Gained by Aluminum Calorimeter + Heat Gained by Water
[tex]m_cC_c\Delta T_c = m_wC_w\Delta T_w + m_aC_a\Delta T_a[/tex]
where,
[tex]m_c[/tex] = mass of copper = 227 g
[tex]m_w[/tex] = mass of water = 844 g
[tex]m_a[/tex] = mass of aluminum = 155 g
[tex]C_c[/tex] = specific heat capacity of calorimeter = 385 J/kg.°C
[tex]C_w[/tex] = specific heat capacity of water = 4200 J/kg.°C
[tex]C_a[/tex] = specific heat capacity of aluminum = 890 J/kg.°C
[tex]\Delta T_c[/tex] = change in temperature of copper = 283°C - T
[tex]\Delta T_w[/tex] = change in temperature of water = T - 14.6°C
[tex]\Delta T_a[/tex] = change in temperature of aluminum = T - 14.6°C
T = equilibrium temperature = ?
Therefore,
[tex](227\ g)(385\ J/kg.^oC)(283^oC-T)=(844\ g)(4200\ J/kg.^oC)(T-14.6^oC)+(155\ g)(890\ J/kg.^oC)(T-14.6^oC)\\\\24732785\ J - (87395\ J/^oC) T = (3544800\ J/^oC) T - 51754080\ J+ (137950\ J/^oC) T-2014070\ J\\\\24732785\ J +51754080\ J+2014070\ J = (3544800\ J/^oC) T+(137950\ J/^oC+(87395\ J/^oC) T\\\\78560935\ J = (3770145\ J/^oC) T\\\\T = \frac{78560935\ J}{3770145\ J/^oC}[/tex]
T = 20.84°C
HELP PLSS I CANT FAIL!!!
Elements from Period 3 of the periodic table are highlighted. Which element
is a metalloid?
A. Sodium
B. Argon
C. Sulfur
D. Silicon
A box is being pulled to the right over a rough surface. t > fk , so the box is speeding up. Suddenly the rope breaks. What happens? The box:_________.
a. keeps its speed for a short while, then slows and stops. slows steadily until it stops.
b. stops immediately.
c. continues speeding up for a short while, then slows and stops.
d. continues with the speed it had when the rope broke.
Answer:
a. keeps its speed for a short while, then slows and stops. slows steadily until it stops.
Explanation:
Since the tension in the rope, t is greater than the kinetic friction fk, the box is moving forward because there is a net force on it. That is, t - fk = f = ma.
Since there is a net force, there is an acceleration and thus an increasing velocity.
When the rope breaks, the tension, t = 0. So, t - fk = 0 - fk = -fk = ma'.
Now, the net force acting on the box is friction in the opposite direction. This force tends to slow the box down from its initial velocity at acceleration, 'a' until its velocity is zero, where it stops. Since the frictional force is constant, the acceleration, a' on the box is thus constant and the box undergoes uniform deceleration until its velocity is zero.
So, the box keeps its speed for a short while, then slows and stops. slows steadily until it stops.
So, the answer is a.
Một vật chuyển động tròn đều có chu kì T = 0,25 s. Tính tần số chuyển động f của vật?
Explanatio
omega=2pi/T
Answer:
0
0000
Explanation:
A reaction requires 1.5 mL of ammonia if it occurs at 1.65 atm and 23 degrees celsius. If the temperature is changed to 30 degrees celsius, what will the new pressure be if the volume remains the same?
Answer: The new pressure will be 1.69 atm
Explanation:
Gay-Lussac's law states that the pressure of the gas is directly proportional to the temperature of the gas at constant volume and the number of moles.
Mathematically,
[tex]P\propto T[/tex] (At constant volume and number of moles)
OR
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex] .....(1)
[tex]P_1\text{ and }T_1[/tex] are the pressure and temperature of the gas
[tex]P_2\text{ and }T_2[/tex] are the final pressure and temperature of the gas
We are given:
[tex]P_1=1.65atm\\T_1=23^oC=[23+273]K=296K\\V_2=?mL\\T_2=30^oC=[30+273]K=303K[/tex]
Putting values in equation 1, we get:
[tex]\frac{1.65atm}{296K}=\frac{P_2}{303K}\\\\P_2=\frac{1.65\times 303}{296}\\\\P_2=1.69atm[/tex]
Hence, the new pressure will be 1.69 atm