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Olegator [25]
2 years ago
6

If a 10 kg ball and a 100 kg ball we’re dropped from a plane, which would hit the ground first?

Physics
1 answer:
meriva2 years ago
3 0

Answer: The 100 kg ball

Explanation:

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sergiy2304 [10]
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Question 1 Gary is on the space shuttle. It takes off and lifts him to a height of 300 km above Earth's surface. a. How has Gary
balandron [24]
A) The mass is an intrinsic property of an object: it means it depends only on the properties of the object, so it does not depend on the location of the object. Therefore, Gary's mass at 300 km above Earth's surface is equal to his mass at the Earth's surface.

b) The weight of an object is given by
W=mg
where
m is the mass
g= \frac{GM}{r^2} is the gravitational acceleration at the location of the object, with G being the gravitational constant, M the mass of the planet and r the distance of the object from the center of the planet.

At the Earth's surface, g=9.81 m/s^2, so Gary's weight is
W=mg=9.81 m  (1)
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Then, we must calculate the value of g at 300 km above Earth's surface. the Earth's radius is 
R=6370 km
So the distance of Gary from the Earth's center is
r=R+h=6370 km+300 km =6670 km = 6.67 \cdot 10^6 m

The Earth's mass is M=5.97 \cdot 10^{24} kg, so the gravitational acceleration is
g'=G \frac{M}{r^2}= (6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )\frac{5.97 \cdot 10^{24} kg}{(6.67 \cdot 10^6 m)^2}=8.95 m/s^2

Therefore, Gary's weight at 300 km above Earth's surface is 
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If we compare (1) and (2), we find that Gary's weight has changed by
\frac{W'}{W}= \frac{8.95 m}{9.81 m}=0.91
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3 years ago
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Goryan [66]
Number one a and b are automatically out if you decrease the pitch or frequency you wont hear it louder it does not add up ok?
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SashulF [63]

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