Answer:
c. V = k Q1 * Q2 / R1 potential energy of Q1 and Q2 separated by R
V2 / V1 = (R1 / R2) = 1/4
V2 = V1 / 4
Answer:
A) 12.57 m
B) 5 RPM
C) 3.142 m/s
Explanation:
A) Distance covered in 1 Revolution:
The formula that gives the relationship between the arc length or distance covered during circular motion to the angle subtended or the revolutions, is given as follows:
s = rθ
where,
s = distance covered = ?
r = radius of circle = 2 m
θ = Angle = 2π radians (For 1 complete Revolution)
Therefore,
s = (2 m)(2π radians)
<u>s = 12.57 m</u>
B) Angular Speed:
The formula for angular speed is given as:
ω = θ/t
where,
ω = angular speed = ?
θ = angular distance covered = 15 revolutions
t = time taken = 3 min
Therefore,
ω = 15 rev/3 min
<u>ω = 5 RPM</u>
C) Linear Speed:
The formula that gives the the linear speed of an object moving in a circular path is given as:
v = rω
where,
v = linear speed = ?
r = radius = 2 m
ω = Angular Speed in rad/s = (15 rev/min)(2π rad/1 rev)(1 min/60 s) = 1.571 rad/s
Therefore,
v = (2 m)(1.571 rad/s)
<u>v = 3.142 m/s</u>
The kinetic molecular theory of matter states that: ... Molecules in the solid phase have the least amount of energy, while gas particles have the greatest amount of energy. The temperature of a substance is a measure of the average kinetic energy of the particles.
Answer:
(a) 4.21 m/s
(b) 24.9 N
Explanation:
(a) Draw a free body diagram of the object when it is at the bottom of the circle. There are two forces on the object: tension force T pulling up and weight force mg pulling down.
Sum the forces in the radial (+y) direction:
∑F = ma
T − mg = m v² / r
v = √(r (T − mg) / m)
v = √(0.676 m (54.7 N − 1.52 kg × 9.8 m/s²) / 1.52 kg)
v = 4.21 m/s
(b) Draw a free body diagram of the object when it is at the top of the circle. There are two forces on the object: tension force T pulling down and weight force mg pulling down.
Sum the forces in the radial (-y) direction:
∑F = ma
T + mg = m v² / r
T = m v² / r − mg
T = (1.52 kg) (4.21 m/s)² / (0.676 m) − (1.52 kg) (9.8 m/s²)
T = 24.9 N