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inna [77]
3 years ago
14

When the mass of an object decreases, the force of gravity

Physics
1 answer:
Viktor [21]3 years ago
8 0
Hello There!

From what i know, gravitational force increases if the mass is increased.
If the mass is being decreased, then i assume it will be B. Decreases.

Hope This Helps You!
Good Luck :) 

- Hannah ❤
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23 and 24 only please help!
Lilit [14]

Answer:

23: Acceleration 24:1m/s^2

Explanation:

a = (v_f - v_i) /delta(t)\\a=(16-10)/6\\a=6/6\\a=1m/s^2

7 0
2 years ago
An aluminium object has a mass of 27kg and density of 2700 the object is atached to a spring and immerged in a tank of water det
laiz [17]

Answer:

0.01m^3

Explanation:

m=27

density=2700

density=mass/volume

2700=27/volume

volume=27/2700=0.01

8 0
2 years ago
A 70.9-kg boy and a 43.2-kg girl, both wearing skates face each other at rest on a skating rink. The boy pushes the girl, sendin
Lelechka [254]

Answer:

Explanation:

Given

mass of boy m_b=70.9\ kg

mass of girl m_g=43.2\ kg

speed of girl after push v_g=4.64\ m/s

Suppose speed of boy after push is v_b

initially momentum of system is zero so final momentum is also zero because momentum is conserved

P_i=P=f

0=m_b\cdot v_b+m_g\cdot v_g

v_b=-\frac{m_g}{m_b}\times v_g

v_b=-\frac{43.2}{70.9}\times 4.64  

v_b=-2.82\ m/s

i.e. velocity of boy is 2.82 m/s towards west                

8 0
3 years ago
Plzzz help will mark the brainliest
ANEK [815]
Ciara is winging....etc
The answer is : 0.60 N, toward the center of the circle


A satellite....etc
The Answer is : 7400 m/s


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8 0
3 years ago
If the charge on the negative plate of the capacitor is 121 nano-Coulomb, how many excess electrons are on that plate? Write you
Julli [10]

Answer:

n = 756.25 giga electrons

Explanation:

It is given that,

If the charge on the negative plate of the capacitor, Q=121\ nC=121\times 10^{-9}\ C

Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :

Q=ne

e is the charge on electron

n=\dfrac{Q}{e}

n=\dfrac{121\times 10^{-9}}{1.6\times 10^{-19}}

n=7.5625\times 10^{11}

or

n = 756.25 giga electrons

So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.

6 0
3 years ago
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