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Blizzard [7]
3 years ago
13

As you walked on the moon, the earth’s gravity would still pull on you weakly and you would still have an earth weight. How larg

e would that earth weight be, compared to your earth weight on the earth’s surface? (Note: The earth’s radius is 6378 km and the distance separating the centers of the earth and moon is 384400 kilometers.)
Physics
1 answer:
vichka [17]3 years ago
6 0

Answer:

g_2 =2\times 10^{-4}g      

Explanation:

given,

radius of earth = 6378 km

distance separating the center of the earth and moon =  384400 km

we know acceleration due to gravity

g = \dfrac{Gm}{r^2}            

g_2 = g\dfrac{r^2}{R^2}                

g_2 = g\dfrac{6378^2}{384400^2}              

g_2 =2\times 10^{-4}g      

hence, the weight of earth compared to the your weight on earths surface is equal to 2 × 10⁻⁴ g.

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How much kinetic energy does a proton gain if it is accelerated, with no friction, through a potential difference of 1.00 V? The
ra1l [238]

Answer:

If energy is conserved, then the sum of the potential energy and the kinetic energy is a constant.

Assuming the proton starts from rest, so it's kineitc energy is zero, but it has a potential energy, PE equal to:

PE = qV

where q =1.6 x 10^-19 C

and V = 1.00 V

Assuming the proton no longer experiences the potential energy and it is all converted to kinetic energy then:

PE* = 0,

KE* = 1/(2mv^2)

Now since

PE + KE = Total energy =PE* + KE*

Therefore,

qV + 0 = 0 + 1/2mv^2

Or

KE = qV = 1.6 10^-19 J

4 0
3 years ago
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sergey [27]
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3 years ago
A 0.2 kg hockey park is sliding along the eyes with an initial velocity of -10 m/s when a player strikes it with his stick, caus
babunello [35]

Answer:

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

Explanation:

The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:

I = m\cdot (\vec{v}_{2} - \vec{v_{1}}) (1)

Where:

I - Impulse, in kilogram-meters per second.

m - Mass, in kilograms.

\vec{v_{1}} - Initial velocity of the hockey park, in meters per second.

\vec{v_{2}} - Final velocity of the hockey park, in meters per second.

If we know that m = 0.2\,kg, \vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right] and \vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right], then the impulse applied by the stick to the park is approximately:

I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right]

I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right]

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

8 0
3 years ago
Define SI unit of measurement.​
ivanzaharov [21]

Answer:

the group of units suggested by the international convention of scientists in 1960 AD to make similarties in meseurment all over the world is called SI units

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3 years ago
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dlinn [17]
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5 0
3 years ago
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