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pychu [463]
3 years ago
13

Waves on a swimming pool propagate at 0.720 m/s. You splash the water at one end of the pool and observe the wave go to the oppo

site end, reflect, and return in 31.0 s. How far away is the other end of the pool?
Physics
1 answer:
SVEN [57.7K]3 years ago
5 0

Answer:

11.16 m

Explanation:

total distance traveled by the wave = speed * time

                                                            = 0.720 m/s * 31.0 s

                                                            = 22.32 m

Since this is an echo back to the starting point then the length of the swimming pool must be half of the distance traveled.

length of swimming pool  =  22.32 m÷2 = 11.16 m

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3 years ago
The question is in the picture
Sedbober [7]

Answer:

e) 120m/s

Explanation:

When the ball reaches its highest point, its velocity becomes zero, meaning

v_0-gt = 0.

where v_0 is the initial velocity.

Solving for t we get

t = \dfrac{v_0}{g}

which is the time it takes the ball to reach the highest point.

Now, after the ball has reached its highest point, it turns around and falls downwards. After time t_0 since it had reached the highest point, the ball has traveled downwards and the velocity v_f it has gained is

v_f = gt_0,

and we are told that this is twice the initial velocity v_0; therefore,

v_f = 2v_0  = gt_0

which gives

t_0 = \dfrac{2v_0}{g}.

Thus, the total time taken to reach velocity 2v_0 is

t_{tot} = t+t_0 = \dfrac{v_0}{g}+\dfrac{2v_0}{g}

t_{tot} = \dfrac{3v_0}{g}.

This t_{tot}, we are told, is 36 seconds; therefore,

36= \dfrac{3v_0}{g},

and solving for v_0 we get:

v_0 = \dfrac{36g}{3}

v_0 = \dfrac{36s(10m/s^2)}{3}

\boxed{v_0 = 120m/s}

which from the options given is choice e.

7 0
4 years ago
If a car travels 600m west in 25 seconds, what is its velocity?
Elena L [17]

v =  \frac{ \:the covered distance}{time} \\

v =  \frac{600}{25} =  \frac{6 \times 100}{25} = 6 \times 4 = 24 \:  \\

v = 24 \:  \frac{m}{s} \\

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If west means the west of the axis x the velocity equal :

- 24 \:  \frac{m}{s} \\

4 0
3 years ago
Which form of energy does flowing water possess
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Answer:

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Explanation:

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2 years ago
Read 2 more answers
A roller coaster car may be approximated by a block of mass m. Thecar, which starts from rest, is released at a height h above t
elena55 [62]

Answer:

The first part can be solved via conservation of energy.

mgh = mg2R + K\\K = mg(h-2R)

For the second part,

the free body diagram of the car should be as follows:

- weight in the downwards direction

- normal force of the track to the car in the downwards direction

The total force should be equal to the centripetal force by Newton's Second Law.

F = ma = \frac{mv^2}{R}\\mg + N = \frac{mv^2}{R}

where N = 0 because we are looking for the case where the car loses contact.

mg = \frac{mv^2}{R}\\v^2 = gR\\v = \sqrt{gR}

Now we know the minimum velocity that the car should have. Using the energy conservation found in the first part, we can calculate the minimum height.

mgh = mg2R + \frac{1}{2}mv^2\\mgh = mg2R + \frac{1}{2}m(gR)\\gh = g2R + \frac{1}{2}gR\\h = 2R + \frac{R}{2}\\h = \frac{5R}{2}

Explanation:

The point that might confuse you in this question is the direction of the normal force at the top of the loop.

We usually use the normal force opposite to the weight. However, normal force is the force that the road exerts on us. Imagine that the car goes through the loop very very fast. Its tires will feel a great amount of normal force, if its velocity is quite high. By the same logic, if its velocity is too low, it might not feel a normal force at all, which means losing contact with the track.

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3 years ago
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