Answer:
F = K Q1 Q2 / R^2 force between 2 charged partices
F2 / F1 = (R1 / R2)^2 = (1 / 2.69)^2 = .139
F2 = .139 F1
Answer:
54.3N
Explanation:
The normal force is perpendicular to the slope, so:
Normal Force = cos(37.2)(9.8*65).......507.39N
F(friction)=mu*F(normal)
F(friction)=(0.107)(507.39)
F(friction)=54.3N
Answer:
A)
0.395 m
B)
2.4 m/s
Explanation:
A)
= mass of the cart = 1.4 kg
= spring constant of the spring = 50 Nm⁻¹
= initial position of spring from equilibrium position = 0.21 m
= initial speed of the cart = 2.0 ms⁻¹
= amplitude of the oscillation = ?
Using conservation of energy
Final spring energy = initial kinetic energy + initial spring energy

B)
= mass of the cart = 1.4 kg
= spring constant of the spring = 50 Nm⁻¹
= amplitude of the oscillation = 0.395 m
= maximum speed at the equilibrium position
Using conservation of energy
Kinetic energy at equilibrium position = maximum spring potential energy at extreme stretch of the spring

Refer to the diagram shown below.
Assume that
(a) The piano rolls down on frictionless wheels,
(b) Wind resistance is negligible.
The distance along the ramp is
d = (1.3 m)/sin(22°) = 3.4703 m
The component of the piano's weight along the ramp is
mg sin(22°)
If the acceleration down the ramp is a, then
ma = mg sin(22°)
a = g sin(22°) = (9.8 m/s²) sin(22°) = 3.671 m/s²
The time, t, to travel down the ramp from rest is given by
(3.4703 m) = 0.5*(3.671 m/s²)*(t s)²
t² = 3.4703/1.8355 = 1.8907
t = 1.375 s
Answer: 1.375 s
Answer:
The car would travel after applying brakes is, d = 14.53 m
Explanation:
Given that,
The time taken to apply brakes fully is, t = 0.5 s
The velocity of the car, v = 29.06 m/s
The distance traveled by the car in 0.5 s, d = ?
The relation between the velocity, displacement, and time is given by the formula
d = v x t m
Substituting the values in the above equation,
d = 29.06 m/s x 0.5 s
= 14.53 m
Therefore, the car would travel after applying brakes is, d = 14.53 m