Okay, haven't done physics in years, let's see if I remember this.
So Coulomb's Law states that

so if we double the charge on

and double the distance to

we plug these into the equation to find
<span>

</span>
So we see the new force is exactly 1/2 of the old force so your answer should be

if I can remember my physics correctly.
Answer: X
Explanation:
This situation can be illustrated as a car in circular motion (image attached).
In circular motion the acceleration vector
is always directed toward the center of the circumference (that's why it's called centripetal acceleration).
So, in this case the arrow labeled X is the only that points toward the center, hence it represents the car's centripetal acceleration
Answer:
fcosθ + Fbcosθ =Wtanθ
Explanation:
Consider the diagram shown in attachment
fx= fcosθ (fx: component of friction force in x-direction ; f: frictional force)
Fbx= Fbcosθ ( Fbx: component of braking force in x-direction ; Fb: braking force)
Wx= Wtanθ (Wx: component of weight in x-direction ; W: Weight of semi)
sum of x-direction forces = 0
fx+ Fbx=Wx
fcosθ + Fbcosθ =Wtanθ
Explanation:
Here's a clearer rendering of the question requirements;
Complete the sentences with the correct wording. Imagine that a force gauge is mounted between the rope and the chain carousel saddle. If you do not touch your feet to the ground when the vehicle is stationary, the dynamometer indicates A / B. When the carousel turns, you will read C / D on the dynamometer.
A. Your weight with the saddle
C. Rope strength value
B. Your weight
D. Centripetal force value
DE which is the differential equation represents the LRC series circuit where
L d²q/dt² + Rdq/dt +I/Cq = E(t) = 150V.
Initial condition is q(t) = 0 and i(0) =0.
To find the charge q(t) by using Laplace transformation by
Substituting known values for DE
L×d²q/dt² +20 ×dq/dt + 1/0.005× q = 150
d²q/dt² +20dq/dt + 200q =150