Answer: 4.5 x 10^-36 N
Explanation: The interaction between the charges is not electrical because the distance ( between the protons) is too large.
The interaction is gravitational because one is placed on earth and the other on the moon.
The force of repulsion between these two protons can be gotten using newton's law of gravitational force which is mathematically given below as
F=(G*m1*m2)/r²
F = Gravitational force
G = gravitational constant =6.674 x 10^-11
m1 = mass of the first proton =1g=0.001kg
m2 = mass of second proton =1g=0.001kg
r = distance between the earth and the moon = 3.844 x 10^9m
Slotting the parameters into the formula, we have that
F= 6.674x10^-11 x 0.001 x 0.001/(3.844 x 10^9)²
F= 6.694 x 10^-17/ 14.77 x 10^18
Finally
F = 4.5 x 10^-36N
Answer:
It will double.
Explanation:
Newton's Law of Gravity states that , where G is the gravitational constant, <em>r</em> is the distance between the objects' centers, and and are the objects' masses. We just have simple math here: by doubling , we double the entire fraction. By doubling the entire fraction, we double the gravitational pull. Therefore, <em>Newton's Law of Gravity states that if the mass of one object doubles, the gravitational pull on a second object will </em><em>double</em><em>.</em>
I hope this helps you understand it! Have a great day, 'kay?
The rotational effect of a force is called torque.it is the cause of rotation or angular deceleration
<span>τ=rXF
where
</span>τ = F r sin @
hope it helps
Answer:
a) p₀ = 1.2 kg m / s, b) p_f = 1.2 kg m / s, c) θ = 12.36, d) v_{2f} = 1.278 m/s
Explanation:
a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved
a) the initial impulse is
p₀ = m v₁₀ + 0
p₀ = 0.6 2
p₀ = 1.2 kg m / s
b) as the system is isolated, the moment is conserved so
p_f = 1.2 kg m / s
we define a reference system where the x-axis coincides with the initial movement of the cue ball
we write the final moment for each axis
X axis
p₀ₓ = 1.2 kg m / s
p_{fx} = m v1f cos 20 + m v2f cos θ
p₀ = p_f
1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ
1.2482 = v_{2f} cos θ
Y axis
p_{oy} = 0
p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ
0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ
0.2736 = v_{2f} sin θ
we write our system of equations
0.2736 = v_{2f} sin θ
1.2482 = v_{2f} cos θ
divide to solve
0.219 = tan θ
θ = tan⁻¹ 0.21919
θ = 12.36
let's look for speed
0.2736 = v_{2f} sin θ
v_{2f} = 0.2736 / sin 12.36
v_{2f} = 1.278 m / s