Answer:
A box sits stationary on a ramp
Explanation:
Static friction is a force which keeps an object at rest as it is in the case of the box. It has to be overcome for the object to be set into motion.
Static force of friction is calculated as follows:
F= μη
F is static force of friction.
μ is the coefficient of static friction.
η is the normal force.
It depends where you are.
-- If you weigh 120 pounds on the Moon,
then your mass is 329.1 kilograms.
-- If you weigh 120 pounds on Mars,
then your mass is 143.8 kilograms.
-- If you weigh 120 pounds on the Earth,
then your mass is 54.4 kilograms.
G = 9.81 m/sec^2) g = 9.81

<span>Solving for velocity : </span>

<span> = 2gh </span>
<span>v = </span>

<span>v = (2 x 9.81 x 10)^1/2 </span>
<span>v = 196.2 m/sec (answer)</span>
Answer:
-0.045 N, they will attract each other
Explanation:
The strength of the electrostatic force exerted on a charge is given by

where
q is the magnitude of the charge
E is the electric field magnitude
In this problem,

(negative because inward)
So the strength of the electrostatic force is

Moreover, the charge will be attracted towards the source of the electric field. In fact, the text says that the electric field points inward: this means that the source charge is negative, so the other charge (which is positive) is attracted towards it.
To solve this problem we will apply the expression of charge per unit of time in a capacitor with a given resistance. Mathematically said expression is given as

Here,
q = Charge
t = Time
R = Resistance
C = Capacitance
When the charge reach its half value it has passed 10ms, then the equation is,




We know that RC is equal to the time constant, then

Therefore the time constant for the process is about 14ms