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icang [17]
3 years ago
9

Gravitational force on the moon is only 1/6 that of the gravitational force on earth. what would be the weight of a 10-kg object

on the moon and on earth? what would be its mass on the moon and on earth?
Physics
1 answer:
MA_775_DIABLO [31]3 years ago
6 0
According to Newton's Second Law of Motion, the force is equal to the mass of an object multiplied by its acceleration. When you talk about gravitational force, the acceleration referred to here is the acceleration due to gravity. This is very familiar to us in physics. The acceleration due to gravity on Earth is equal to 9.81 m/s². It actually depends on the location. According to the Universal Law of Gravitation:

F = Gm₁m₂/d²

The force is a factor of the product of two masses and their distance from each other. The G is a constant called the universal gravitational constants. So, gravitational force is actually a relative force exerted by one body to another. 

Going back the Second Law of Motion, we can modify the equation to:

F = mg

Since it is mentioned that the gravity on the moon is only 1/6 of the Earth, then the gravity for moon is:
g,moon = 1/6(9.81) = 1.635 m/s²
So, let's compare the weight of the object with a mass of 10 kg. The weight is actually the force due to gravity pulling you towards the center of the body.

Weight on Earth = (10 kg)(9.81 m/s²) = 98.1 N
Weight on Moon = (10 kg)(1.635 m/s²) = 16.35 N

The mass, on the other hand, is not affected by gravity. It is always constant. Therefore, the mass of the object on the moon is the same with its mass on the Earth.
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A 40-kg rock is dropped from an elevation of 10 m. What is the velocity of the rock when it is 5-m from the ground?
ivolga24 [154]

Answer:

Explanation:

Given

mass of rock m=40\ kg

Elevation of Rock h=10\ m

Distance traveled by rock with time

h=ut+\frac{1}{2}at^2

where, u=initial velocity

t=time

a=acceleration

here initial velocity is zero

when rock is 5 m from ground then it has traveled a distance of 5 m from top because total height is 10 m

5=0\times t+\frac{1}{2}(9.8)(t^2)

t^2=\frac{10}{9.8}

t=1.004\approx 1\ s

velocity at this time

v=u+at

v=0+9.8\times 1.004

v=9.83\ m/s

6 0
3 years ago
If points a and b are connected by a wire with negligible resistance, find the magnitude of the current in the 12.0 v battery.
Marizza181 [45]
V = I * R
Where V is the voltage, I is the current and R is the resistance. Using Ohm's law, you require resistance to find the current through the wire. Technically, if the wire has a resistance of 0, you will get infinite current. But this isn't possible. Maybe the negligible resistance refers to the battery's internal resistance - not the wire's resistance. 
7 0
3 years ago
Sound is faster than light that is why we hear thunder before we see lightning
Umnica [9.8K]

Answer:

false

Explanation:

sound travels slower than light. that is why we see lightning before we hear the thunder

8 0
3 years ago
A ball is thrown vertically upwards with a velocity
zhuklara [117]

Answer:

Explanation:

The acceleration of gravity is 9.8m/s^2.

So to calculate the time it will take to make the ball stop(which btw means the ball now reach its greatest height), use the formula V1=V0+at. V1 is the final velocity(which is 0), V0 is the starting velocity(which is 30m/s), and the a(cceleration) is 9.8m/s^2.

(You can ignore the fact "at" is -30 instead 30, it's because the directions two velocity travel are opposite. )

We can now know the time it takes to make the ball stop just by the gravitational force is about 3 sec.

Use another formula S=1/2at^2, to find out the S(height) is 1/2*9.8*3^2=44.1, which is approximately D.45m .

6 0
2 years ago
A car passes point “A” and then 120 meters later. It’s velocity was measured 21 m/s. If it’s acceleration was constant at 0.853
Norma-Jean [14]

Recall that

{v_f}^2-{v_i}^2=2a\Delta x

where v_i and v_f are the initial and final velocities, respecitvely; a is the acceleration; and \Delta x is the change in position.

So we have

\left(21\dfrac{\rm m}{\rm s}\right)^2-{v_i}^2=2\left(0.853\dfrac{\rm m}{\mathrm s^2}\right)(120\,\mathrm m)

\implies v_i\approx\boxed{15.4\dfrac{\rm m}{\rm s}}

(Normally, this equation has two solutions, but we omit the negative one because the car is moving in one direction.)

7 0
2 years ago
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