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kati45 [8]
3 years ago
11

A stone is dropped into water from a bridge 52 m above the water's

Physics
1 answer:
dusya [7]3 years ago
3 0

Answer:

Its final velocity and how much time it takes to reach the water

Explanation:

The motion of the stone is a uniformly accelerated motion, so we can use the following suvat equation to determine its final velocity:

v^2-u^2=2as

where

v is the final velocity

u = 0 is the initial velocity

a=g=9.8 m/s^2 is the acceleration of gravity

s = 52 m is the distance covered during the fall

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0^2+2(9.8)(52)}=31.9 m/s

We can also find how much time it takes to reach the water, using the equation

v=u+at

where

v = 31.9 m/s is the final velocity

u = 0 is the initial velocity

a=g=9.8 m/s^2

t is the time

And solving for t,

t=\frac{v-u}{a}=\frac{31.9-0}{9.8}=3.26 s

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Imagine a ringing bell set inside a sealed glass jar. Once all the air is removed and a vacuum is crated, the ringing sound is n
kenny6666 [7]

working...

Sound wave needs medium to travel

as energy which travels in this wave is because of transfer from one particle to another particle

If there is no medium then energy can not be transferred and sound wave will not travel

so in vacuum we can not listen sound

similarly here air is removed it means there is no medium inside the jar to travel the sound and hence we can not hear it

Option B is correct

Without air, the sound waves cannot travel to the ear.

5 0
3 years ago
Which runner has greater kinetic energy: a 45 kg runner moving at a speed of 7 m per second or a 93 kg runner moving at a
bagirrra123 [75]

Kinetic Energy = (1/2) (mass) (speed)

First runner:  KE = (1/2) (45kg) (49 m/s)  =  1,102.5 Joules  

Second runner:  KE = (1/2) (93kg) (9 m/s)  =  418.5 Joules

The <em>first runner </em><em>has 163</em>% more kinetic energy than the second runner has.

7 0
2 years ago
A runner is moving at a constant speed of 8.00 m/s around a circular track. If the distance from the runner to the center of the
Genrish500 [490]

Answer: Last option

2.27 m/s2

Explanation:

As the runner is running at a constant speed then the only acceleration present in the movement is the centripetal acceleration.

If we call a_c to the centripetal acceleration then, by definition

a_c =w^2r = \frac{v^2}{r}

in this case we know the speed of the runner

v =8.00\ m/s

The radius "r" will be the distance from the runner to the center of the track

r = 28.2\ m

a_c = \frac{8^2}{28.2}\ m/s^2

a_c = 2.27\ m/s^2

The answer is the last option

3 0
3 years ago
Luis wants to compare the density of three solid objects that are different shapes and sizes. What information does he need to m
tatyana61 [14]

Answer:

u need to make sure that comparison is = to shapes and then find the shapes sizes and add them

7 0
3 years ago
Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w
gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

u(x) = \frac{-2}{x} + \frac{4}{x^2}

At x = 1.00 m

u(1) = \frac{-2}{1} + \frac{4}{1^2}

= 4J

Kinetic energy = (1/2)mv²

= \frac{1}{2} * 4(3)^2

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

u(5) = \frac{-2}{5} + \frac{4}{5^2}

= \frac{4-10}{25} = \frac{-6}{25} J

= -0.24J

Kinetic energy =

\frac{1}{2} * 4Vf^2

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

Vf = \sqrt{frac{22.4}{2}

= 3.33 m/s

b) magnitude of force when x = 5.0m

u(x) = \frac{-2}{x} + \frac{4}{x^2}

\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}

= \frac{2}{x^2} - \frac{8}{x^3}

At x = 5.0 m

\frac{2}{5^2} - \frac{8}{5^3}

F = \frac{2}{25} - \frac{8}{125}

= 0.016N

8 0
3 years ago
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