PHYSICS COLLEGE

Dogs can hear higher-pitched whistles that humans do. How do you
think the sound frequencies that dogs can
hear compare to the frequencies that humans
can hear?

Answers

Answer 1
Answer:

Dogs can hear sounds at higher frequencies than humans. The range of sound frequencies that dogs can hear is approximately 40 Hz to 60,000 Hz, while the range for humans is 20 Hz to 20,000 Hz. This means that dogs can hear ultrasonic sounds that are beyond the range of human hearing.

What is sound about?

In terms of physics, sound is a vibration that travels through a transmission medium like a gas, liquid, or solid as an acoustic wave.

Sound is the reception of these waves and the brain's perception of them in terms of human physiology and psychology. Dogs have the ability to hear ultrasonic sounds that are audible only to them.

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A suspended platform of negligible mass is connected to the floor below by a long vertical spring of force constant 1200 N/m. A circus performer of mass 70 kg falls from rest onto the platform from a height of 5.8 m above it. Find the maximum spring compression

Answers

Answer:

The maximum spring compression = 3.21 m

Explanation:

The height of the circus performer above the platform connected to string material = 5.8 m

Let the maximum compression of the spring from the impact of the circus performer be x.

According to the law of conservation of energy, the difference in potential energy of the circus performer between the initial height and the level at which spring is compressed to is equal to the work done on the spring to compress it by x

Workdone on the spring by the circus performer = (1/2)kx²

where k = spring constant = 1200 N/m

Workdone on the spring by the circus performer = (1/2)(1200)x² = 600x²

The change in potential energy of the circus performer = mg (5.8 + x)

m = mass of the circus performer = 70 kg

g = acceleration due to gravity = 9.8 m/s²

The change in potential energy of the circus performer = (70)(9.8)(5.8 + x) = (3978.8 + 686x)

600x² = 3978.8 + 686x

600x² - 686x - 3978.8 = 0

Solving this quadratic equation

x = 3.21 m or - 2.07 m

Since the negative answer doesn't satisfy the laws of physics, our correct answer is 3.21 m

Hope this Helps!!!

What is the frequency of a photon that has the same momentum as a neutron moving with a speed of 1.90 × 103 m/s?

Answers

The mass of a neutron is:
m=1.67 \cdot 10^(-27)kg
Since we know its speed, we can calculate the neutron's momentum:
p=mv=(1.67 \cdot 10^(-27)kg)(1.90 \cdot 10^3 m/s)=3.17 \cdot 10^(-24) kg m/s

The problem says the photon has the same momentum of the neutron, p.  The photon momentum is given by
p= (h)/(\lambda)
where h is the Planck constant, and \lambda is the photon wavelength. If we re-arrange the equation and we use the momentum we found before, we can calculate the photon's wavelength:
\lambda= (h)/(p)= (6.6 \cdot 10^(-34)Js)/(3.17 \cdot 10^(-24) kg m/s)=2.08 \cdot 10^(-10) m

And since we know the photon travels at speed of light c, we can now calculate the photon frequency:
f= (c)/(\lambda)= (3 \cdot 10^8 m/s)/(2.08 \cdot 10^(-10) m)= 1.44 \cdot 10^(18) Hz

N5
Swinging a tennis racket against a ball is an example of a third class lever.
OT
OF
9
Please select the best answer from the choices provided.
K

Answers

Final answer:

Swinging a tennis racket against a ball as a third class lever in physics.


Explanation:

Swinging a Tennis Racket as a Third Class Lever

A tennis racket swinging against a ball is indeed an example of a third class lever in physics. In a third class lever, the effort is situated between the fulcrum and the load. In this case, the effort is provided by the player's hand gripping the racket handle, the fulcrum is the wrist joint, and the load is the ball being struck by the racket.

When a player swings the racket, the force applied by the player's hand exerts an effort on the handle of the racket. This causes the racket to rotate about the wrist joint acting as the fulcrum. The ball serves as the load, receiving the force and accelerating in the opposite direction.


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Susan’s 10 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30° above the floor. The tension is a constant 30 N and the coefficient of friction is 0.20. Use work and energy to find Paul’s speed after being pulled 3.0 m.

Answers

Final answer:

In this problem, we have calculated the work done by Susan pulling her baby brother on a mat and the work done against friction. The net work done, which is the work done by Susan's pulling minus the work done against friction, is transformed into kinetic energy, giving us the baby's speed after being pulled 3m, which is approximately 1.95 m/s.

Explanation:

To answer this question, we first need to calculate the work done by Susan when she pulls the mat over the distance of 3.0 meters. The angle at which the rope is pulled does make a difference in this calculation. The force that is actually contributing to the work is the horizontal component of the tension, which can be determined by the equation Fh = F cos θ which equals 30N * cos30 = 25.98N.

The work done, W, is equal to this force multiplied by the displacement, so W = Fd = 25.98N * 3m = 77.94 Joules.

Next, we need to calculate the work done against friction. The force of friction is calculated as Ff = µN. Here N is the normal force, which is equal to the weight of the baby, so N = mg = 10kg * 9.8m/s² = 98N. The force of friction then is Ff = µN = 0.20 * 98N = 19.6N. The work done against friction is Wf = Ff * d = 19.6N * 3m = 58.8 Joules.

The net work done on the baby is the work done by Susan minus the work done against friction, so Wnet = W - Wf = 77.94J - 58.8J = 19.14 Joules. This net work is equal to the change in kinetic energy of the baby, ∆K, since Kinitial = 0 (Paul starts at rest), the work done is all transformed into final kinetic energy. So ∆K = 19.14J.

The kinetic energy of an object is given by the equation K = 1/2 mv², so we have 19.14J = 1/2 * 10kg * v². Solving for v gives us roughly v = 1.95 m/s. Therefore, the speed of the baby after being pulled 3 meters is approximately 1.95 m/s.

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Final answer:

To determine Paul's speed, we must calculate the net work done on him using the work-energy theorem. This includes the work done by Susan and the work done against friction. Paul’s speed after being pulled 3.0 m is approximately 1.96 m/s.

Explanation:

Solving this problem involves understanding the work-energy theorem and forces. First, let's calculate the work done. The work done by the force Susan applies (W1) is the product of the tension (T), the distance (d), and the cosine of the angle (θ). W1 = T * d * cos(θ) = 30N * 3.0m * cos(30) = 77.94J.

Next, the work done against friction (W2) is the product of the frictional force and the distance, which is µmgd. Here, µ is the coefficient of friction (0.20), m (10kg) is the mass of the baby, g (9.8m/s2) is the acceleration due to gravity, and d is the distance (3.0 m). W2 = µmgd = 0.20 * 10kg * 9.8m/s2 * 3.0m = 58.8J.

According to the work-energy theorem, the net work done on an object is equal to the change in its kinetic energy. Therefore, the final kinetic energy (and thus the final speed) of Paul will be the initial kinetic energy plus the net work done on him. His initial speed is assumed to be zero, hence the initial kinetic energy is zero. The net work done on him is W = W1 - W2= 77.94J - 58.8J = 19.14J. Setting this equal to the final kinetic energy, (1/2)mv2, allows us to solve for the final speed, v = sqrt((2 * W)/m) = sqrt((2 * 19.14J)/10kg) = 1.96 m/s approximately.

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The electric field at the distance of 3.5 meters from an infinite wall of charges is 125 N/C. What is the magnitude of the electric field 1.5 meters from the wall?

Answers

Explanation:

It is given that,

Distance, r = 3.5 m

Electric field due to an infinite wall of charges, E = 125 N/C

We need to find the electric field 1.5 meters from the wall, r' = 1.5 m. Let it is equal to E'. For an infinite wall of charge the electric field is given by :

E=(\lambda)/(2\pi \epsilon_o r)

It is clear that the electric field is inversely proportional to the distance. So,

(E)/(E')=(r')/(r)

E'=(Er)/(r')

E'=(125* 3.5)/(1.5)  

E' = 291.67 N/C

So, the magnitude of the electric field 1.5 meters from the wall is 291.67 N/C. Hence, this is the required solution.

Since fusion and fission are opposite processes that both produce energy,why can we not simply run the process forward and then backwardrepeatedly and have a limitless supply of energy?A. The products of a fission reaction cannot be used for a fusionreaction, and the products of a fusion reaction cannot be used fora fission reaction.B. Fusion reactions can occur cheaply enough, but fission requiresvery high temperatures.C. Fusion produces energy from nuclei larger than iron, and fissionproduces energy from nuclei smaller than iron.D. Fission reactions can occur cheaply enough, but fusion requires very high temperatures

Answers

ANSWER:

D. Fission reactions can occur cheaply enough, but fusion requires very high temperatures

STEP-BY-STEP EXPLANATION:

One of the main reasons fusion power cannot be harnessed is that its power requirements are incredibly high. For fusion to occur, a temperature of at least 100,000,000°C is needed.

Therefore, the correct answer is D. Fission reactions can occur cheaply enough, but fusion requires very high temperatures
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