Answer:
h = 7.54 m
t = 1.24 s
Explanation:
1.Let g = 9.81 m/s2 is the gravitational acceleration. Since the formula for potential energy is:
where m = 0.25 is the ball mass and h is the height. We can solve for h
2. The time it take for the ball to reach a distance of 7.54m with a gravitational acceleration of 9.81m/s2:
Answer:
1, 1583.33 V/m
2, 4.72*10^-12 C
3, 39.2*10^-11 C
Explanation:
1
E = V / d
E = 8.55 / 5.4*10^-3
E = 8.55 / 0.0054
E = 1583.33 V/m
2
Capacitance, C = (k * e0 * A) / d, where k = 1
A = area of capacitor, 3.37 cm² = 3.37*10^-4 m²
d = plate separation, 5.4 mm
e0 = Constant, 8.85*10^-12
Applying these, we have
C = (1 * 8.85*10^-12 * 3.37*10^-4) / 5.4*10^-3
C = 29.82*10^-16 / 0.0054
C = 5.52*10^-13 F
Since Q = CV, then
Q = 5.52*10^-13 * 8.55
Q = 4.72*10^-12 C
3
We are given that k = 83, so
Capacitance, C = (k * e0 * A) / d
C = (83 * 8.85*10^-12 * 3.37*10^-4) / 5.4*10^-3
C = 2.475*10^-13 / 0.0054
C = 4.58*10^-11 F
Q = CV
Q = 4.58*10^-11 * 8.55
Q = 39.2*10^-11 C
Answer:
Vf = 28 m/s
Explanation:
In order to find the final velocity of the rock, we will use the 3rd equation of motion. The third equation of motion for vertical direction is written as follows:
2gh = Vf² - Vi²
where,
g = acceleration due to gravity = 9.8 m/s²
h = height dropped = 40 m
Vf = final velocity of the rock = ?
Vi = Initial Velocity of the rock = 0 m/s (since, rock was initially at rest)
Therefore,
(2)(9.8 m/s²)(40 m) = Vf² - (0 m/s)²
Vf = √(784 m²/s²)
<u>Vf = 28 m/s</u>
Well since the bowling ball is at rest and is not moving at all, and assuming that the ball is on an even surface, the Fg is equal to the FN = normal force, and thus all forces are balanced, the Fnet = 0.
I say it is false that is the correct answer
