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Luden [163]
3 years ago
10

A mass of 5kg accelerates at 3m/s/s, how much force was put on it?

Physics
2 answers:
Diano4ka-milaya [45]3 years ago
7 0

F = m a

The force is (5kg)x(3 m/s^2)= 15 Newtons, and it's still there.

If the force stops, the acceleration stops.

goldfiish [28.3K]3 years ago
5 0

Answer:

15N

Explanation:

According to Newton's Second Law of Motion

F = m*a

mass = m = 5Kg

acceleration = a = 3m/s^2

=> F = 5kg * 3m/s^2

=> F = 15 N

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Hello! You can call me Emac or Eric.

I understand your problem, that question is pretty hard. But I found some information that I think you should read. This can get your problem done quickly.

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Here is some good information that could help you out a lot!


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The ancient Greek recommendation of rest in the treatment of abnormal behavior soon gave rise to ______. A. Trephining B. The as
Lilit [14]
C. Is the correct answer
7 0
3 years ago
A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.
iris [78.8K]

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

F = \frac{kq_1q_2}{d^2}

Here

k = Coulomb's Constant

q_{1,2} = Charge of each object

d = Distance

Our values are given as,

q_1 = 1 \mu C

q_2 = 6 \mu C

d = 1 m

k =  9*10^9 Nm^2/C^2

a) The electric force on charge q_2 is

F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}

F_{12} = 54 mN

Force is positive i.e. repulsive

b) As the force exerted on q_2 will be equal to that act on q_1,

F_{21} = F_{12}

F_{21} = 54 mN

Force is positive i.e. repulsive

c) If q_2 = -6 \mu C, a negative sign will be introduced into the expression above i.e.

F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}

F_{12} = F_{21} = -54 mN

Force is negative i.e. attractive

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3 years ago
1. What affects a material's resistance?
Ghella [55]

Answer:

A. Thickness and temperature

Explanation:

3 0
3 years ago
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