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NARA [144]
4 years ago
6

A gymnast is swinging on a high bar. The distance between his waist and the bar is 0.829 m, as the drawing shows. At the top of

the swing his speed is momentarily zero. Ignoring friction and treating the gymnast as if all of his mass is located at his waist, find his speed at the bottom of the swing.
Physics
1 answer:
vivado [14]4 years ago
6 0

To find the speed with the given values we can apply the energy conservation equations for which we have that the increase in potential energy is compensated in the decrease of the kinetic energy or vice versa.

Since there is conservation and part of the balance we have to

KE = PE

Where,

KE = Kinetic Energy

PE = Potential Energy

The values for this energy are given as

\frac{1}{2}mv^2 = mgh

\frac{1}{2} v^2 = gh

v = \sqrt{2gh}

v = \sqrt{2(9.8)(0.829)}

v = 4.03m/s

Therefore the speed at the bottom of the swing is 4.03m/s

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Calculate the magnitude of the normal force on a 25.2 kg block in the following circumstances. (Enter your answers in N.) HINT (
irga5000 [103]

Answer:

when the body is resting N = 246.96 N

when the body is resting on a tilted surface N = 212.12 N.

when the body is in a elevator N = 317.036 N

Explanation:

when the block is resting on a stationary surface the normal force is balanced by the weight of the body.

weight of the body = mg = 25.2×9.8 = 246.96 N

therefore normal force = 246.98 N.

when the block is resting on a tilted surface the normal force will be balanced by the \cos \theta component of the weight where Ф is the angle of inclination.

therefore N = mg\cos \Phi

                N= 212.12 N.

when the block is resting on a elevator that is accelerated upward the normal force will be the sum of weight and force due to acceleration ma

therefore N = 246.98 + 25.2×2.78

N = 317.036 N

5 0
4 years ago
You drop a 30 g pebble down a well. You hear a splash 2.7 s later. Ignoring air resistance, how deep is the well? Assume g = 9.8
lys-0071 [83]
I will assume here that the well is sufficiently short so that the time the sound takes to come from the bottom of the well to our ear is negligible.

Since the pebble moves by uniformly accelerated motion, the distance it covers is given by
S= \frac{1}{2}gt^2
where 
g=9.81 m/s^2 is the gravitational acceleration
t=2.7 s is the time the pebble takes to reach the bottom of the well

Therefore, the depth of the well is
S= \frac{1}{2}(9.81 m/s^2)(2.7 s)^2 = 35.7 m \sim 36 m
and the correct answer is B.
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7 0
3 years ago
a car tire is filled with air at pressure 325000 pa at 283 k. if the tire warms up to 302 k, what is the new pressure of the tir
WARRIOR [948]

Answer:

346819 Pa       or   ,347000 Pa in 3 significant figures

Explanation:

P1= 325000Pa , T1= 283K ,

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as here volume and mass both are constant so using ratio method for pressure temperature law we have P1/T1 = P2/T2

THIS WE CAN ALSO OBTAIN BY RATIO METHOD FOR GENERAL GAS LAW AS

P1V1/(m1T1 ) =  P2 V2/ (m2 T2)

IF V1 = V2 =V AND m1=m2=m then expression reduces to

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P2=346819

P2 = 347000 Pa in 3 significant figures

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