Answer:
Explanation:
As the disc is unrolling from the thread then at any moment of the time
We have force equation as
also by torque equation we can say
Now we have
Also from above equation the tension force in the string is
Answer:
A. 5.4 * 10^(-4) m
B. 500V
Explanation:
A. Electric potential, V is given as:
V = kq/r
This means that radius, r is
r = kq/V
r = (9 * 10^9 * 30 * 10^(-12))/500
r = (270 * 10^(-3))/500
r = 5.4 * 10^(-4) m
B. Now the radius is doubled and the charge is doubled,
V = (9 * 10^9 * 2 * 30 * 10^(-12))/(2 * 5.4 * 10^(-4) * 2)
V = 500V
Answer:
The ball will have an upward velocity of 6 m/s at a height of 5.51 m.
Explanation:
Hi there!
The equations of height and velocity of the ball are the following:
y = y0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
y = height at time t.
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).
v = velocity of the ball at time t.
Placing the origin at the throwing point, y0 = 0.
Let´s use the equation of velocity to obtain the time at which the velocity is 12.0 m/s / 2 = 6.00 m/s.
v = v0 + g · t
6.00 m/s = 12.0 m/s -9.81 m/s² · t
(6.00 - 12.0)m/s / -9.81 m/s² = t
t = 0.612 s
Now, let´s calculate the height of the baseball at that time:
y = y0 + v0 · t + 1/2 · g · t² (y0 = 0)
y = 12.0 m/s · 0.612 s - 1/2 · 9.81 m/s² · (0.612 s)²
y = 5.51 m
The ball will have an upward velocity of 6 m/s at a height of 5.51 m.
Have a nice day!
