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Fudgin [204]
3 years ago
10

Why is the gravitational potential energy of an object 1 meter above the moon’s surface less than its potential energy 1 meter a

bove Earth’s surface?
A. The object’s mass is less on the moon.
B. The object’s weight is more on the moon.
C. The object’s acceleration due to gravity is less on the moon.
D. both a and c
Physics
1 answer:
dsp733 years ago
3 0

Potential energy relative to the surface is

         (mass) x (<span>acceleration due to gravity) x (height above the surface).

At 1.0 meter above the surface, that is

</span>          (mass) x (<span>acceleration due to gravity) x (1.0 meter) .

The object's mass doesn't change, so the only thing that has any effect
on its potential energy at 1 meter above the surface is the acceleration
of gravity or, in other words, the surface of <em><u>what</u></em> ?

</span>
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Answer:

the can's kinetic energy is 0.42 J

Explanation:

given information:

Mass, m = 460 g = 0.46 kg

diameter, d = 6 cm, so r = d/2 = 6/2 = 3 cm = 0.03 m

velocity, v = 1.1 m/s

the kinetic energy of the can is the total of kinetic energy of the translation and rotational.

KE = \frac{1}{2} I ω^2 + \frac{1}{2} mv^{2}

where

I = \frac{1}{2} mr^{2} and ω = \frac{v}{r}

thus,

KE = \frac{1}{2} \frac{1}{2} mr^{2} (\frac{v}{r})^2 + \frac{1}{2} mv^{2}

     = \frac{1}{2} \frac{1}{2} mr^{2} \frac{v^{2} }{r^{2}} + \frac{1}{2} mv^{2}

     = \frac{1}{4} mv^{2} + \frac{1}{2} mv^{2}

     = \frac{3}{4} mv^{2}

     = \frac{3}{4} (0.46) (1.1)^{2}

     = 0.42 J

8 0
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What happens to the atomic number of an atom when the number of neutrons in the nucleus of that atom increases? a It decreases b
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Answer:

It remains the same

Explanation:

It remains the same. This is because the number of protons doesn't change and the number of protons determines the atomic number.

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gayaneshka [121]
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A boy who is riding his bicycle, moves with an initial velocity of 5 m/s. ten second later, he is moving at 15 m/s. what is his
pantera1 [17]

\Large {{ \sf {Question :}}}

<h3>A boy who is riding his bicycle, moves with an initial velocity of 5 m/s. Ten second later, he is moving at 15 m/s. What is his acceleration?</h3>

\Large {{ \sf {Given :}}}

<h3>Initial Velocity (<em>u</em>) - 5 m/s</h3><h3>Final Velocity (<em>v</em>) - 15 m/s</h3><h3>Time (<em>t</em>) - 10 sec</h3>

\Large {{ \sf {Formulae  :}}}

<h3>If the velocity of an object changes from an initial value <em>u </em>to the final value <em>v </em>in time <em>t,</em><em> </em>the acceleration <em>a</em> is, </h3><h3>a \:  =  \frac{v - u}{t}</h3><h3>\Large {{ \sf {Step-by-step explanation :}}}</h3>

a \:  =  \frac{v - u}{t}  \\ or \:  \: a =  \frac{(15 - 5)}{10} m \: s^{ - 2}  \\ or \:  \: a \:  =  \frac{10}{10}m \: s^{ - 2} \\ or \:  \: a = 1m \: s^{ - 2}

\Large {{ \sf {Answer :}}}

<h3>His acceleration is </h3><h3>1m \: s^{ - 2}</h3><h3 /><h3 />
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