Answer:
The force of static friction acting on the luggage is, Fₓ = 180.32 N
Explanation:
Given data,
The mass of the luggage, m = 23 kg
You pulled the luggage with a force of, F = 77 N
The coefficient of static friction of luggage and floor, μₓ = 0.8
The formula for static frictional force is,
Fₓ = μₓ · η
Where,
η - normal force acting on the luggage 'mg'
Substituting the values in the above equation,
Fₓ = 0.8 x 23 x 9.8
= 180.32 N
Hence, the minimum force require to pull the luggage is, Fₓ = 180.32 N
1) Force = m*a = 1.00 g * (1kg / 1000 g) * 225 m/s^2 = 0.225 N
2) Charge
Force = K (charge)^2 /(distance)^2 => charge = √ [Force * distance^2 / k]
k = 9.00 * 10^9 N*m^2 / C^2
charge = √ [0.225 N * (0.02 m)^2 / 9.00* 10^9 N*m^2 / C^2 ]
charge = 0.0000001 C = 0.0001 mili C
Answer:
F = 768 N
Explanation:
It is given that,
Speed of the elevator, v = 3.2 m/s
Grain drops into the car at the rate of 240 kg/min, 
We need to find the magnitude of force needed to keep the car moving constant speed. The relation between the momentum and the force is given by :
Since, the speed is constant,
F = 768 N
So, the magnitude of force need to keep the car is 768 N. Hence, this is the required solution.
Answer:
0 J
Explanation:
Kinetic energy is defined as:
KE = 1/2 m v²
where m is mass and v is velocity.
The car starts at rest, so it has zero velocity. Therefore, its initial kinetic energy is 0 J.
