The object takes 0.5 seconds to complete one rotation, so its rotational speed is 1/0.5 rot/s = 2 rot/s.
Convert this to linear speed; for each rotation, the object travels a distance equal to the circumference of its path, or 2<em>π</em> (1.2 m) = 2.4<em>π</em> m ≈ 7.5 m, so that
2 rot/s = (2 rot/s) • (2.4<em>π</em> m/rot) = 4.8<em>π</em> m/s ≈ 15 m/s
thus giving it a centripetal acceleration of
<em>a</em> = (4.8<em>π</em> m/s)² / (1.2 m) ≈ 190 m/s².
Then the tension in the rope is
<em>T</em> = (50 kg) <em>a</em> ≈ 9500 N.
Answer:
E = 16.464 J
Explanation:
Given that,
Mass of tetherball, m = 0.8 kg
It is hit by a child and rises 2.1 m above the ground, h = 21. m
We need to find the maximum gravitational potential energy of the ball. The formula for the gravitational potential energy is given by :
E = mgh
g is acceleration due to gravity
E = 0.8 kg × 9.8 m/s² × 2.1 m
= 16.464 J
So, the maximum potential energy of the ball is 16.464 J.
Answer:
Straight line in the direction of the tangential velocity the ball had at the moment the string broke
Explanation:
After the string breaks, the ball now disconnected from the centripetal force that was exerted via the string, continues its travel in a straight line in the direction of the tangential velocity it had at the moment the string broke.
Bohr’s model showed that electron orbits had distinct radii.
Answer:
v = 1.98*10^8 m/s
Explanation:
Given:
- Rod at rest in S' frame
- makes an angle Q = sin^-1 (3/5) in reference frame S'
- makes an angle of 45 degree in frame S
Find:
What must be the value of v if as measured in S the rod is at a 45 degree)
Solution:
- In reference frame S'
x' component = L*cos(Q)
y' component = L*sin(Q)
- Apply length contraction to convert projected S' frame lengths to S frame:
x component = L*cos(Q) / γ (Length contraction)
y component = L*sin(Q) (No motion)
- If the rod is at angle 45° to the x axis, as measured in F, then the x and y components must be equal:
L*sin(Q) = L*cos(Q) / γ
Given: γ = c / sqrt(c^2 - v^2)
c / sqrt(c^2 - v^2) = cot(Q)
1 - (v/c)^2 = tan(Q)
v = c*sqrt( 1 - tan^2 (Q))
For the case when Q = sin^-1 (3/5)::
tan(Q) = 3/4
v = c*sqrt( 1 - (3/4)^2)
v = c*sqrt(7) / 4 = 1.98*10^8 m/s
