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3 years ago

Number of waves that pass a given point in one second

Physics

1 answer:

Studentka2010 [4]3 years ago

5 0

number of waves that pass a given point in one second is called frequency..

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PLEASE HELP ME I DONT UNDERSTAND THIS AND THIS IS TIMED

mixer [17]

(C)

Explanation:

The circle has a radius r = 0.5 m, which means that its circumference C is

$C = 2\pi r = 2\pi(0.5\:\text{m}) = 3.14\:\text{m}$

One revolution means that the stopper travels a distance equal to the circumference of the circle so the velocity of the stopper is

$v = \dfrac{C}{t} =\dfrac{3.14\:\text{m}}{0.2\:\text{s}} = 15.7\:\text{m/s} \approx 16\:\text{m/s}$

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2 years ago

Danny Diver weighs 500 N and steps off a diving board 10 m above the water. Danny hits the water with kinetic energy of

Morgarella [4.7K]

Answer:

Danny hits the water with kinetic energy of 5000 J.

Explanation:

Given that,

The Weight of Danny Diver,

F = 500 N

m*g= 500 N

He steps off a diving board 10 m above the water.

h=10 m

when Danny diver hits water he generates the kinetic energy.

We need to find the kinetic energy of the water.

Let kinetic energy is K.

K = m*g*h

Where g is acceleration due to gravity.

that g= 9.8 m/s^2

now substituting the values in above equation

K= (500) * 10

K= 5000 J

Hence,

he hits the water with kinetic energy of 5000 J.

Learn more about Kinetic energy here:

brainly.com/question/15587458

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7 0

2 years ago

Look at the divisions between 2 and 4. What level of precision does this stopwatch have based on the divisions marked on its fac

klemol [59]

Answer:

10ths of a second

Explanation:

Because I just did it

8 0

3 years ago

6. The hole on a level, elevated golf green is a horizontal distance of 150 m from the tee and at an elevation of 12.4 m above t

Georgia [21]

Answer:

u = 104.68 m/s

Explanation:

given,

horizontal distance = 150 m

elevation of 12.4 m

angle = 8.6°

horizontal motion = x = u cos θ. t .............(1)

vertical motion =

$y = u sin \theta - \dfrac{1}{2}gt^2$ ................(2)

from equation(1) and (2)

$y = x tan \theta - \dfrac{gx^2}{2u^2cos^2\theta}$ ..........{3}

$12.4 = 150\times tan (8.6) - \dfrac{9.8\times 150^2}{2u^2cos^2(8.6)}$

$\dfrac{9.8\times 150^2}{2u^2cos^2(8.6)} = 10.29$

$\dfrac{9.8\times 150^2}{2\times 10.29\times cos^2(8.6)} = u^2$

$u = \sqrt{10959.34}$

u = 104.68 m/s

The initial speed of the ball is u = 104.68 m/s

8 0

3 years ago

A brick with a mass of 20 kg and a weight of 196 N is released on a frictionless plane at an angle of 30°. What is the accelerat

Veronika [31]

Answer:4.9m/s^2

Basically answer A

Explanation:

6 0

3 years ago