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Y_Kistochka [10]
3 years ago
13

A 38.0 kg child is in a swing that is attached to ropes 1.70 m long. The acceleration of gravity is 9.81 m/s 2 . Find the gravit

ational potential energy associated with the child relative to the child’s lowest position under the following conditions:
Physics
1 answer:
siniylev [52]3 years ago
7 0

Answer:

at zero point : GPE = 0 J

at max height : GPE = 633.7 J

Explanation:

the gravitational potential energy at the lowest point is zero

maximum height relative to the lowest point = h =1.70 m

G potential energy at max height = mgh = (38kg)(9.81m/s^2)(1.7)

                                           = 633.7 J

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At its lowest setting a centrifuge rotates with an angular speed of ω1 = 250 rad/s. When it is switched to the next higher setti
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Answer:

Part(a): The angular acceleration is 5.63~rad~s^{-2}.

Part(b): The angular displacement is 2629~rad.

Explanation:

Part(a):

If \omega_{1},~\omega_{2}~and~\alpha be the initial angular speed, final angular speed and angular acceleration  of the centrifuge respectively, then from rotational kinematic equation, we can write

\alpha = \dfrac{\omega_{2} - \omega_{1}}{t}......................................................(I)

where 't' is the time taken by the centrifuge to increase its angular speed.

Given, \omega_{i} = 250~rad~s^{-1}, \omega_{f} = 750~rad~s^{-1} and t = 9.5~s. From equation (I), the angular acceleration is given by

\alpha = \dfrac{750 - 250}{9.5}~rad~s^{-2} = 5.63~rad~s^{-2}

Part(b):

Also the angular displacement (\Delta \theta) can be written as

&&\Delta \theta = \omega_{1}~t + \dfrac{1}{2}\alpha~t^{2}\\&or,& \Delta \theta = (250 \times 9.5 + \dfrac{1}{2} \times 5.63 \times 9.5^{2})~rad = 2629~rad

8 0
3 years ago
How to get a + b on a graph
Nataly [62]

The first positively essential requirement is that
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3 0
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1. A student adds water and sugar to a jar and seals the jar so that nothing can get in or out. The
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Answer:

The mass will stay the same throughout time

3 0
3 years ago
How many squares for three traits?
Studentka2010 [4]
You just need one square but with 16 little boxes 
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4 0
3 years ago
a street light is mounted at the top of a 15 foot pole. A man 6 ft tall walks away from the pole wit a speed of 7 ft/s along a s
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Answer:

16.3 ft/s

Explanation:

Let d=distance

and

x = length of shadow.

Therfore,

x=(d + x)

 = 6/15

So,

    15x = 6x + 6d

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x = (2/3)d.

As we know that:

dx=dt

   = (2/3) (d/dt) 

Also,

Given:

d(d)=dt

     = 7 ft/s

Thus,

d(d + x)=dt

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Substitute, d= 7  

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