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Margarita [4]
3 years ago
10

Sound waves travel at the rate of 343 m/s at 20°C. If a man standing 450 meters away from the wall of a canyon yells, “Hello,” h

ow long will it take for him to hear his own voice echo off of the wall?
Physics
1 answer:
Klio2033 [76]3 years ago
7 0

Answer:

2.62seconds

Explanation:

Echo occurs when sound reflects and causing a delay before getting to the listener.

Let x be the distance from between the source and the reflector

t be the time taken for the echo to occur

v be the velocity of sound in air.

Using the relationship 2x = vt

t = 2x/v

Given x = 450metres

v = 343m/s

t = 2(450)/343

t = 900/343

t = 2.62seconds

It will take 2.62 seconds for him to hear his own voice echo off of the wall.

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Answer:

Energy transferred = 28.8 Joules.

1. Energy transferred = 144 Joules.

2. The unit of potential difference, volts can also be described as Joules per Coulombs.

3. Current, I = 6.945 Amperes.  

Explanation:

<u>Part A.</u>

Given the following data;

Current, I = 1.2A

Time, t = 2 minutes

Potential difference, V = 12 volts.

To find the energy transfered;

Energy transferred = charge moved * potential difference

E = Q * V

Substituting into the equation, we have;

Energy transferred = (1.2 * 2) * 12

Energy transferred = 2.4 * 12

Energy transferred = 28.8 Joules.

<u>Part B.</u>

1. <em><u>Given the following data;</u></em>

Charge, Q = 24C

Potential difference = 6V

To find the energy transferred;

E = Q * V

Substituting into the equation, we have;

E = 24 * 6

E = 144 Joules.

2. Since we know that, Energy transferred = charge moved * potential difference

Potential \; difference = \frac {Energy \; transferred}{Charged \; moved}

The units of energy is Joules while the unit of the quantity of charge moved is Coulomb.

Therefore, the unit of potential difference becomes Joules per Coulomb.

3. <em><u>Given the following data;</u></em>

Potential difference = 18V

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To find the current;

E = Q * V

Substituting into the equation, we have;

500 = Q*18

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Q = 27.78C

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Answer:

v_o = 4.54 m/s  

Explanation:

<u>Knowns  </u>

From equation, the work done on an object by a constant force F is given by:  

W = (F cos Ф)S                                   (1)  

Where S is the displacement and Ф is the angle between the force and the displacement.  

From equation, the kinetic energy of an object of mass m moving with velocity v is given by:  

K.E=1/2m*v^2                                       (2)

From The work- energy theorem , the net work done W on an object equals the difference between the initial and the find kinetic energy of that object:  

W = K.E_f-K.E_o                                 (3)

<u>Given </u>

The displacement that the sled undergoes before coming to rest is s = 11.0 m and the coefficient of the kinetic friction between the sled and the snow is μ_k = 0.020  

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We know that the kinetic friction force is given by:

f_k=μ_k*N

And we can get the normal force N by applying Newton's second law to the sled along the vertical direction, where there is no acceleration along this direction, so we get:  

∑F_y=N-mg

     N=mg

Thus, the kinetic friction force is:  

f_k = μ_k*N  

Since the friction force is always acting in the opposite direction to the motion, the angle between the force and the displacement is Ф = 180°.  

Now, we substitute f_k and Ф into equation (1), so we get the work done by the friction force:  

W_f=(f_k*cos(180) s

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Since the sled eventually comes to rest, K.E_f= 0 So, from equation (3), the net work done on the sled is:  

W= -K.E_o    

Since the kinetic friction force is the only force acting on the sled, so the net work on the sled is that of the kinetic friction force  

W_f= -K.E_o  

From equation (2), the work done by the friction force in terms of the initial speed is:  

W_f=-1/2m*v^2  

Now, we substitute for W_f= -μ_k*mg*s, and solving for v_o so we get:  

-μ_k*mg*s = -1/2m*v^2  

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Finally, we plug our values for s and μ_k, so we get:  

v_o = √2 x (0.020) x (9.8 m/s^2) x (11.0 m) = 4.54 m/s  

v_o = 4.54 m/s  

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