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Darina [25.2K]
3 years ago
10

A swimmer standing near the edge of a lake notices a cork bobbing in the water. While watching for one minute, she notices the c

ork bob (from up to down to back up) 240 times. What is the frequency in Hz of the water wave going by?
Physics
1 answer:
timurjin [86]3 years ago
5 0

Answer:

4Hz (240 cycles/60 seconds = 4 cycles/second)

Explanation:

hope this helped!

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A proton is moving toward a second, stationary proton. What happens as the protons get closer?
MatroZZZ [7]

Answer:

A. Kinetic energy is converted to electric potential energy, and the proton moves more slowly.

Explanation:

When a moving proton is brought close to a stationary one, the kinetic energy of the moving one is converted to electric potential  and the proton moves more slowly.

Kinetic energy is the energy due to the motion of a body. A moving proton will possess this form of energy.

Two protons according to coulombs law will repel each other with an electrostatic force because they both have similar charges. This will increase their electric potential energy of both of them.

Potential energy is the energy at rest of a body. As it increases, the motion of a body will be slower and it will tend towards being stationary.

5 0
3 years ago
A machine can never be 100% efficient because some work is always lost due to .
klasskru [66]
Friction
Hope it helped
5 0
3 years ago
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قوة الجذب المركزي تكون في اتجاه
pochemuha

Answer:

تكون دائمًا متعامدة مع سرعة الجسم وتكون دائمًا في اتجاه مركز انحناء المسار

Explanation:

6 0
3 years ago
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If the amplitude of a wave is increased the frequency of the wave will
Travka [436]

Answer:

Remain the same

Explanation:

There is no relationship between amplitude frequency.

3 0
3 years ago
If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of laun
zvonat [6]

Answer:

H = 1/2 g t^2    where t is time to fall a height H

H = 1/8 g T^2   where T is total time in air  (2 t  = T)

R = V T cos θ       horizontal range

3/4 g T^2 = V T cos θ       6 H = R    given in problem

cos θ = 3 g T / (4 V)           (I)

Now t = V sin θ / g     time for projectile to fall from max height

T = 2 V sin θ / g

T / V = 2 sin θ / g

cos θ = 3 g / 4 (T / V)     from (I)

cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ

tan θ = 2/3      

θ = 33.7 deg

As a check- let V = 100 m/s

Vx = 100 cos 33.7 = 83,2

Vy = 100 sin 33,7 = 55.5

T = 2 * 55.5 / 9.8 = 11.3 sec

H = 1/2 * 9.8 * (11.3 / 2)^2 = 156

R = 83.2 * 11.3 = 932

R / H = 932 / 156 = 5.97        6 within rounding

3 0
2 years ago
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