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Lerok [7]
3 years ago
14

Find the height of a baseball with a mass of 0.15 kg that has a GPE of 73.5 J. Help please?

Physics
2 answers:
LUCKY_DIMON [66]3 years ago
8 0
Hight is 50 b/c it is on the ledt side



Slav-nsk [51]3 years ago
4 0
Gravitational Potential Energy of an object, relative to the ground =

           (mass) x (acceleration of gravity) x (height above the ground)


               73.5 J  =  (0.15 kg)  x  (9.8 m/s²)  x  (height above the ground).

1 J = 1 kg · m² / s²

               73.5 kg·m²/s²  =  (0.15 kg) · (9.8 m/s²) · (height)

Divide each side by (0.15 kg): 

               (73.5 kg·m² / (0.15 kg·s²)      = (9.8 m/s²) · (height)

Divide each side by (9.8 m/s²):

               (73.5 kg·m²·s²) / (0.15 kg·s² · 9.8 m)  =  (height)

After doing all the arithmetic on the left side:

               (73.5) / (0.15 · 9.8) (m)   =   height

                       50                 (m)    =   height
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A 100-W lightbulb is placed in a cylinder equipped with a moveable piston. The lightbulb is turned on for 0.010 hour, and the as
Taya2010 [7]

Answer:

w =  - 508.53 joules

q = - 3091.47 joules

Explanation:

Let us convert the time in hours into seconds

0.010* 3600\\= 36

Change in internal energy

\delta E = p * \delta t

where E is the internal energy in Joules

p is the power in watts

and t is the time in seconds

\delta E = - 100 * 36\\

\delta E = - 3600 Joules

Amount of work done by the system

w = - P * \delta V

where P is the pressure and V is the volume

Substituting the given values in above equation, we get -

w = - 1 * ( 5.92 -0.90)\\

w = -5.02 liter-atmospheres

Work done in Joules

- 5.02 * 101.3\\= 508.53Joules

q = \delta E - w\\

Substituting the given values we get -

q = - 3600 - (-508.53)\\q = - 3091.47

Thus

w =  - 508.53 joules

q = - 3091.47 joules

7 0
3 years ago
Write the ratio of the following?<br><br>CaCO3<br>C2H6<br>Fe(NO3)3​
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Answer:

C2H6 up the road to be with its own in

4 0
2 years ago
Why does a third class lever cannot magnify force?​
maw [93]

Explanation:

The third class lever cannot magnify our force because in third class lever the effort it between the load and the fulcrum. Also, in this type of lever no matter where the force is applied, it is always greater than the force of load. Hence, That type of lever cannot magnify our force.

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3 years ago
The slope of a distance-time graph will give
Dafna1 [17]
5.
explanation: the answer is 5
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A gyroscope slows from an initial rate of 32.0 rad/s at a rate of 0.700 rad/s2. How long does it take to come to rest? How many
Darya [45]

Answer:

t=45.7s

\alpha =116revolutions

Explanation:

Since we have given values of ω₀=32.o rad/s ,ω=0 and α=-0.700 rad/s² to find t we use below equation

w=w_{o}+at\\  0=(32.0rad/s)+(-0.700rad/s^{2} )t\\t=\frac{-32.0}{-0.700} \\t=45.7s

To find revolutions we use below equation

w^{2}=w_{o}^{2}+2a\alpha

Substitute the given values to find revolutions α

So

0=(32.0rad/s)^{2}+2(-0.700rad/s^{2} )\alpha  \\\alpha =\frac{(-32.0rad/s)^{2}}{2(-0.700rad/s^{2} )} \\\alpha =731rad

To convert rad to rev:

\alpha =(731rad)*(\frac{1rev}{2\pi rad} )\\\alpha =116revolutions

7 0
3 years ago
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