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Mademuasel [1]
3 years ago
13

B. Calculate the rock's gravitational potential energy at 50 m, 20 m, 1 m, and om high. Put the

Physics
1 answer:
klasskru [66]3 years ago
4 0

The gravitational potential energy of the rock is:

490 J at 50 m

196 J at 20 m

9.8 J at 1 m

Explanation:

The gravitational potential energy (GPE) of an object is the energy possessed by the object due to its position in the gravitational field.

For an object near the Earth's surface, the GPE is given by

GPE=mgh

where

m is the mass of the object

g=9.8 m/s^2 is the acceleration of gravity at Earth's surface

h is the heigth of the object relative to the ground

For the rock in  this problem,

m = 1 kg

So its GPE at h = 50 m is

GPE(50)=(1)(9.8)(50)=490 J

The GPE at h = 20 m is

GPE(20)=(1)(9.8)(20)=196 J

While the GPE at h = 1 m is

GPE(1)=(1)(9.8)(1)=9.8 J

Learn more about gravitational potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

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A contestant in a winter games event pulls a 36.0 kg block of ice across a frozen lake with a rope over his shoulder as shown in
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(a) The minimum force F he must exert to get the block moving is 38.9 N.

(b) The acceleration of the block is 0.79 m/s².

<h3>Minimum force to be applied </h3>

The minimum force F he must exert to get the block moving is calculated as follows;

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Fcosθ = μ(s)mg

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The vapor pressure of benzene, C6H6, is 40.1 mmHg at 7.6°C. What is its vapor pressure at 60.6°C? The molar heat of vaporization
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Answer:

The vapor pressure at 60.6°C is 330.89 mmHg

Explanation:

Applying Clausius Clapeyron Equation

ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}[\frac{1}{T_1}- \frac{1}{T_2}]

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P₁ is the initial vapor pressure of benzene = 40.1 mmHg

T₂ is the final temperature of benzene = 60.6°C = 333.6 K

T₁ is the initial temperature of benzene = 7.6°C = 280.6 K

ΔH is the molar heat of vaporization of benzene = 31.0 kJ/mol

R is gas rate = 8.314 J/mol.k

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