Answer:
A closed organ pipe is λ/4 (node-antinode) long.
λ = speed / frequency = 331.5 / 259.6 = 1.28 m
λ/4 = .319 m length of closed pipe
An open pipe has a fundamental wavelength of A-N-A or λ/2
The second harmonic would be A-N-A-N-A or λ = 1.28 m for the second harmonic 331.5 / 259.6 = 1.28 (the fundamental would be 331.5 / .628
A 3.1 kg ball is dropped from the top of a 38 m tall building. What is the speed of the ball when it is halfway from the building to the ground? Round your answer to 2 decimal places.
Answer:
19.3m/s
Explanation:
Use third equation of motion
[tex]v^2-u^2=2gh[/tex]
where v is the velocity at halfway, u is the initial velocity, g is gravity (9.81m/s^2) and h is the height at which you'd want to find the velocity
insert values to get answer
[tex]v^2-0^2=2(9.81m/s^2)(38/2)\\v^2=9.81m/s^2 *38\\v^2=372.78\\v=\sqrt[]{372.78} \\v=19.3m/s[/tex]
The potential at point P is the work required to bring a one-coulomb test charge from far
away to the point P?
True or false ?
A first order reaction is 25% completed in 1h minutes. Calculate the time required for its 50% completion.
At the end of meiosis II, the new haploid cells in humans are known as which of the following?
Diploid cells
Embryo
Zygotes
Gametes
compare and contrast a transverse wave and a compressional wave Give an example for each type
Answer:
Transverse waves oscillate perpendicular to the direction of the wave (e.g. any electromagnetic wave like radiowaves, x-rays...) whilst compressional waves oscillate in the same direction of the wave (e.g. sound waves)
Explanation:
Which is better, forward bias or reverse bias, and why ?!
Answer:
reverse bias
Explanation:
bcz the potential barrier and impedes the flow of charge carriers. In contrast, a forward bias weakens the potential barrier, thus allowing current to flow more easily across the junction.
If a rocket experiences an acceleration generated by the gravity force between the earth and itself, what is this acceleration if the rocket flies 1000 km above the ground and the Earth's radius is 6.378 * 10 ^ 6 * r m. We know the Earth has a mass of 5.97*10^ 24 kg(in m/s^ 2 , G=6.67*10^ -11 N(m/kg)^ 2 ) ?
a 8.97
b 7.32
c 9.81
d 5.5
e 11.45
This question involves the concepts of Gravitational Force and Weight force.
The value of acceleration is "b. 7.32 m/s²".
At the given height the weight of the rocket must be equal to the gravitational force between rocket and the Earth:
[tex]W=F_G\\mg=\frac{GmM}{R^2}\\\\g=\frac{GM}{R^2}[/tex]
where,
g = acceleration = ?
G = universal gravitational constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = mass of earth = 5.97 x 10²⁴ kg
R = Radius of Earth + Height = 6.378 x 10⁶ m + 1 x 10⁶ m = 7.378 x 10⁶ m
Therefore,
[tex]g=\frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}{(7.378\ x\ 10^6\ m)^2}[/tex]
g = 7.32 m/s²
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when do we say a curvilinear motion is accelerated?
Answer:
There always is an acceleration in a curvilinear motion, as the velocity vector changes, so always.
True or False. Father of a multitude" obeyed God's command to kill His son because the Ten Commandments had not yet been given.
Answer:
Explanation:
He agreed to the command, because it was God who proposed it. The story is supposed to make a comment on faith. Abraham had faith that somehow God would make everything right. The Ten commandments came later, but really had nothing to do with Abraham's decision.
True.
Two moles of an ideal monatomic gas go through the cycle abc. For the complete cycle, 800 J of heat flows out of the gas. Process ab is at constant pressure, and process bc is at constant volume. States a and b have temperatures Ta = 200 K and Tb = 300 K.
(a) Sketch the all possible pV-diagrarns for the cycle.
(b) What is the work W for the process ca?
a) Sketches of all possible pv-diagrams for the cycle are attached below
b) The work W[tex]_{ac}[/tex] for the process Ca is : 2462.8 J
Given data :
Amount of heat flowing out = 800 J
Ta = 200 K
Tb = 300 K
R = 800
B) Determine the work W for the process Ca
Wₐs = -pdv
= - [ pVb - pVa ] ---- ( 1 )
note : pVb = nRTb , pVa = nRTa
Equation ( 1 ) becomes
= -nR [ Tb - Ta ]
= - 2(8.314 ) [ 300 - 200 ]
= - 1662.87
given that W[tex]_{bs}[/tex] = 0 which is isochonic
dv = 0 ( cyclic process ) = d∅ - dw
∴ 0 = 800 - ( Wₐs + W[tex]_{ac}[/tex] )
Therefore : W[tex]_{ac}[/tex] = 800 + 1662.8 = 2462.8 J
Hence we can conclude that the work W for the process Ca = 2462.8 J
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True or False. Isaac at first resisted his father's command to be the sacrifice.
when a student comned his dry hair for a long time, his hair began to stand up. Which of these has MOST LIKELY happened
His hair got charged due to continuous friction causing them to get attracted to other objects and thus standing up.
Answer:The comb and hair have become charged with opposite charges.
Explanation:When a comb is run through your hair charges pass between your hair and the comb, so the comb becomes charged either positively or negatively, and the hair oppositely charged. When the comb is brought close to paper an opposite charge is induced in the paper, and the opposite charges attract.
Two positive charges of 1 mC and 10 mC are separated by a distance of 10 m. Find the
direction and the magnitude of electrostatic force between the charges. |Describe the
direction in terms of the charges attract each other," or "the charges repel each other."
mili C=1043 C
Answer:
900N, charges repel
Explanation:
F = KQq/d²
K = 9 × 10^9Nm²/C²
Q= 10mC = 10 × 10^-3C
q = 1mc = 1 × 10^-3C
d = 10m
F = ?
Force = (9 × 10^9 × 10 × 10^-3 × 1 × 10^-3)/10²
= 9 × 10²
= 900N
It will be an electrostatic force of repulsion since like charges(two positive charges) repel
The magnitude and the direction of the electrostatic force between the charges will be 900N and the charges will repel each other.
What is Coulomb's law of forces?Coulomb's law states that whenever the two charged particles are separated by a particular distance then there will be a force of attraction or repulsion acts between the charge.
The formula of coulomb's force will be given by
[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]
[tex]K=9\times 10^9\ \frac{Nm^2}{C^2}[/tex]
[tex]q_1= 10\times 10^{-3}\ C[/tex]
[tex]q_2=1\times 10^{-3}\ C[/tex]
[tex]\rm Distance \ d=10 \ m[/tex]
Now the force will be calculated as
[tex]F=(9\times 10^9)\dfrac{(10\times 10^{-3})\times (1\times 10^{-3})}{10}[/tex]
[tex]F= 900\ N[/tex]
The charges are positive in nature so they will repel each other
Hence the magnitude and the direction of the electrostatic force between the charges will be 900N and the charges will repel each other.
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How much energy is needed to change the temperature of 50g of water 15°c
Explanation:
This question is not feasible. There is no way to calculate the energy needed because the question is missing the final temperature
A 1.2 kg hammer slams down on a nail at 5.0 m/s and bounces off at 1.0 m/s. If the impact lasts 1.0 ms, what average force is exerted on the nail?
Answer:
Explanation:
Impulse results in a change of momentum
FΔt = mΔV
F = mΔV/Δt
The impulse acting on the hammer will equal the impulse acting on the nail
If we assume upward is the positive direction
F = m(vf - vi)/t
F = 1.2(1.0 - (-1.5)) / 0.001
F = 3000 N
A 1.2 kg hammer slams down on a nail at 5.0 m/s and bounces off at 1.0 m/s, if the impact lasts 1.0 ms, so the average force exerted on the nail is 3000 N.
What is average force?According to Newton's third law, there is an equal and opposite force for every force. The hammer's force on the nail will be identical in size to the nail's force on the hammer, but opposite in direction.
Momentum shifts as a result of impulse.
FΔt = mΔV
F = mΔV/Δt
The force exerted on the hammer will be equivalent to the force exerted on the nail.
m(vf - vi)/t = F
F = 1.2(1.0 - (-1.5)) / 0.001
F = 3000 N
Therefore, F = 3000 N, if A 1.2 kg hammer slams down on a nail at 5.0 m/s and bounces off at 1.0 m/s, if the impact lasts 1.0 ms.
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The magnetic field B at all points within the colored circle of the figure (Figure 1) has an initial magnitude of 0.780 T. (The circle could represent approximately the space inside a long, thin solenoid.) The magnetic field is directed into the plane of the diagram and is decreasing at the rate of 0.0300 T/s.
a) What is the magnitude of the induced electric field at any point on the circular conducting ring with radius r = 0.100 m ?
b) What is the direction of this field at any point on the circular conducting ring?
c) What is the current in the ring if its resistance is 4.00 Ω ?
d) What is the emf between points a and b on the ring?
e) If the ring is cut at some point and the ends are separated slightly, what will be the emf between the ends?
The magnitude of the induced electrical field is 0.0015V/m, the field is pointing towards the clockwise direction while the current in the ring will be 0.0002355A if the resistance is 4 ohms. The emf between point a and b is zero and the EMF across the point if they're slightly separated between the ends is 0.000942V
To solve this question, we would have to go about each one individually
Data:
[tex]r=10cm=0.1m\\[/tex]
a.
The magnitude of the induced electrical field at any point within the radius is
[tex]\int\limits^a_b {E} \, du=\frac{dU}{dt}=\pi \frac{dB}{dt}=\pi r^{2}\frac{dB}{dt}\\E*2\pi r=\pi r^{2}\frac{dB}{dt} \\E=\frac{r}{2}\frac{dB}{dt}=\frac{0.1}{2}*0.03=0.0015V/m[/tex]
b.
The field is pointing towards the clockwise direction.
c.
The current in the ring if we are given a resistance of 4ohms
[tex]I=\frac{emf}{R}=\frac{\pi r^{2}\frac{dB}{dt} }{R} =\frac{\pi (0.1)^2*0.03}{4} =0.0002355A[/tex]
d.
The emf between point a and b is zero
e.
The EMF across two points if they're separated by small distance across the ring is
we would use the formula to solve for the EMF
[tex]E=\pi r^{2}\frac{dB}{dt}=\pi (0.1)^2*0.03=0.000942V[/tex]
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The amplitude of a lightly damped oscillator decreases by 4.2% during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle?
Answer:
V= A ω maximum KE of object in SHM
V2 / V1 = .958 ratio of amplitudes since ω is constant
KE2 / KE1 = 1/2 m V2^2 / (1/2 m V1^2) = (V2 / V1)^2
KE2 / KE1 = .958^2 = .918
So KE2 = .918 KE1 and .082 = 8.2% of the energy is lost in one cycle
which alkali metal is most reactive
An ammeter has azero offset error This fault will affect
Answer:
An ammeter has a zero offset error. This fault will affect. neither the precision nor the accuracy of the readings. only the precision of the readings.
The accuracy of the present measurement will be impacted by an ammeter's zero offset inaccuracy, but the precision of the measurement will not change.
What is resistance?Resistance is the obstruction of electrons in an electrically conducting material.
voltage = current ×resistance
How accurate measurement or reading is may be determined by its accuracy. The measurements will be off because of zero inaccuracy.
Thus, The precision of the measurement would not alter if an ammeter made a mistake with its zero offsets, but the accuracy of the current measurement would be decreased.
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objects want to ______ ___________ doing what they're __________ ____________ because they are "lazy." This is called __________.
Answer:
Explanation:
Objects want to continue doing what they're already doing because they are "lazy." This is called inertia.
the driver of a 2.0 × 10³ kg red car traveling on the highway at 45m/s slams on his brakes to avoid striking a second yellow car in front of him, which had come to of because locking ahead.After the brakes are applied a constant friction force of 7.5 × 10³ N acts on the car ignore air resistance.
a₎ Determine the least distance should the brakes be applied to avoid a collision with the other vehicle?
Answer:
Explanation:
a = F/m = 7500/2000 = 3.75 m/s²
v² = u² + 2as
s = (v² - u²) / 2a
s = (0² - 45²) / (2(-3.75))
s = 270 m
a
A person throws a ball up into the air, and the ball falls back towar
would the kinetic energy be the lowest? (1 point)
at a point before the ball hits the ground
when the ball leaves the person's hand
o when the ball is at its highest point
at a point when the ball is still rising
Answer:
when the ball is at its highest point
Explanation:
Provided the ball returns to where it was thrown. The velocity, and therefore kinetic energy, will be momentarily zero at the highest point of the throw.
Please help me answer the following question!
What is the momentum of a 750-kg Volkswagen Beetle when at rest?
Please give the Value and Unit
Answer:
P = 0
Explanation:
Momentum is defined as the quentity of motion contained in a body. Mathematically, it can be defined as the product of mass and velocity. So, in order to determine the Volkswagen Beetle, at rest, we can use the simple formula, as follows:
where,
P = Momentum of Volkswagen Beetle = ?
m = Mass of Volkswagen Beetle = 750 kg
v = Velocity of Volkswagen Beetle = 0 m/s (since, it is at rest)
Therefore,
P =
A runner of mass 80 kg is moving at 8.0 m/s. Calculate her kinetic energy.
Answer:
2560J
Explanation:
By definition the kinetic energy can be calculated in the following way:
K = (mv²)/2 = 80kg·(8.0m/s)²/2 = 2560 J
how is the atomic mass determined?
Answer:
Atomic mass is defined as the number of protons and neutrons in an atom, where each proton and neutron has a mass of approximately 1 amu (1.0073 and 1.0087, respectively). The electrons within an atom are so miniscule compared to protons and neutrons that their mass is negligible.
Explanation:
A rifle is aimed horizontally at a target 47 m away. The bullet hits the target 2.3 cm below the aim point.
Answer:
Is your question asking for the muzzle velocity of the bullet?
Explanation:
I will assume it does
The bullet travels horizontally to the target in the same amount of time it falls 2.3 cm from vertical rest
s = ½at²
t = √(2s/g) = √(2(0.023) / 9.8) = 0.0685118...s
v = d/t = 47/0.0685118 = 686.01242...
v = 690 m/s
1. A roller coaster with a mass of 800 kg sits stationary at the top of a section of track, 75 m above
the ground as shown. When the brake is released, it starts to roll down the track
2. For each height represented in the diagram, calculate the gravitational potential energy using
Ep = mgh. Show ONE SAMPLE calculation in the calculations section below and fill in Table 1 for
each of the heights of the roller coaster. (6 marks)
3. Assuming there is no friction, determine the mechanical kinetic energy using Ek = Etotal - Ep.
Show ONE SAMPLE calculation in the calculations section below and fill in Table 1 for each of
the heights of the roller coaster. (6 marks)
4. For each height represented in the diagram, calculate the velocity using = �2
. Show ONE
SAMPLE calculation in the calculations section below and fill in Table 1 for each of the heights of
the roller coaster. (6 marks)
5. Use your answers to graph how gravitational potential energy, mechanical kinetic energy, and
velocity change as the roller coaster changes height. Use different colours for the three lines on
the graph. Graph paper is provided below. (3 marks)
6. Repeat steps 1 – 5 above for a roller coaster cart that has a mass of 300 kg and enter your
results in Table 2.
Calculations:
800 kg roller coaster cart:
Sample calculation for gravitational potential energy:
Sample calculation for Mechanical kinetic energy:
Sample calculation for velocity:
300 kg roller coaster cart:
Sample calculation for gravitational potential energy:
Sample calculation for mechanical kinetic energy:
Sample calculation for velocity:
Results:
Table 1: Potential energy, kinetic energy, total energy, and velocity of the 800 kg roller coaster cart
Table 2: Potential energy, kinetic energy, total energy, velocity of the 300 kg roller coaster cart.
Graphs:
It’s graphing time. These graphs are a bit different than the ones you did in the
data analysis assignment at the beginning of the course. In this case you have
three things to graph on each graph. (One graph for the 800 kg roller coaster cart
and one graph for the 300 kg roller coaster cart.) You need to graph the
gravitational potential energy with respect to height, the mechanical kinetic
energy vs height, and the velocity vs height.
Let’s look at the energy graphs first. In this case both kinetic energy and
mechanical energy cover the same range of values. This means they can use the
same scale on the y-axis. So, you will use the left y-axis and the x-axis to graph
the kinetic energy vs height and the potential energy vs height. You will need a
legend to explain which line is which. Colour coding is a nice way to highlight this.
The velocity values are much different than the energy values. This means you
need a totally different scale. So, your left y-axis won’t work. You need to make a
second scale on the right y-axis for your velocity values. You will plot the points
the same way as normal, but you will use the numbers on the right-hand scale
instead. Again, be sure to add your velocity line to the legend with a separate
colour code.
Discussion Questions:
1. Describe the relationship between the gravitational potential energy and the mechanical kinetic
energy of the roller coaster on your graph. (2 marks)
2. Describe the shapes of each of the three lines in the graph. Explain why the velocity is different.
(4 marks)
3. Describe how mass affects the speed at the bottom of the roller coaster. (2 marks)
4. Describe how mass affects the gravitational potential energy and the mechanical kinetic energy
of the roller coaster. (2 marks)
5. At what point does the roller coaster have a maximum value for the following? Justify your
answer by explaining why. (2 marks each)
a. Gravitational potential energy
b. Mechanical energy
c. Velocity
6. In your calculations, you assumed that the roller coaster was frictionless. All real roller coasters
encounter friction. Describe how the actual values of the variables would differ, or not differ,
from your calculated values for a real roller coaster. (Hint: what form of energy would some of
the total energy be converted to if there was friction in the system?) (4 marks)
How you will be graded:
Grades will be based on answering questions to demonstrate an understanding of the material covered
in this unit. Point form answers are okay if ideas are complete and use vocabulary (Word Bank)
provided. For questions out of 4 marks, there are 4 responses expected.
Answer:
Give me some hint please
Based on the calculations, potential energy of this roller coaster at a height of 75 meters is equal to 588,000 Joules.
How to calculate potential energy?Mathematically, potential energy is calculated by using this formula:
P.E = mgh
Where:
P.E represents potential energy.m is the mass.h is the height.g is acceleration due to gravity.Note: Acceleration due to gravity is equal to 9.8 m/s².
At a height of 75 m, we have:
P.E = 800 × 9.8 × 75
P.E = 588,000 Joules.
At a height of 60 m, we have:
P.E = 800 × 9.8 × 60
P.E = 470,400 Joules.
At a height of 45 m, we have:
P.E = 800 × 9.8 × 45
P.E = 352,800 Joules.
At a height of 30 m, we have:
P.E = 800 × 9.8 × 30
P.E = 235,200 Joules.
At a height of 15 m, we have:
P.E = 800 × 9.8 × 15
P.E = 117,600 Joules.
In conclusion, we can deduce that the potential energy of this roller coaster decreases with a decrease in height.
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please help me !!
Gymnasts often practice on foam floors, which increase the collision time when a gymnast falls. What effect does this have on collisions?
A. The change in momentum needed to stop the gymnast is increased.
B. The change in momentum needed to stop the gymnast is decreased.
C. The force exerted by the floor on the gymnast decreases.
D. The force exerted by the floor on the gymnast increases.
how long is the photons journey from the milky way to earth
A 4.0 kilogram projectile is fired horizontally from a 500 kilogram cannon initially at rest. The momentum of the projectile after being fired is 600 kilogram-meters per second to the north
(neglecting friction).
What is the speed of the cannon after firing?
0.83 m/s
1.2 m/s
o
3.3 m/s
150 m/s
The speed of the cannon after firing is 1.2 m/s
This can be solved using the law of conservation of momentum.
From the law of conservation of momentum,
⇒ The momentum of the projectile is equal to the momentum of the cannon.
MV = P................ Equation 1⇒ Where :
M = mass of the cannon,V = velocity of the cannonP = momentum of the projectile.⇒ make V the subject of the equation
V = P/M.................. Equation 2From the question,
⇒ Given:
P = 600 kgm/sM = 500 kg⇒ Substitute these values into equation 2
V = 600/500V = 1.2 m/sHence, The speed of the cannon after firing is 1.2 m/s
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