Answer:
Explanation:
angular momentum of the putty about the point of rotation
= mvR where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .
= .045 x 4.23 x 2/3 x .95 cos46
= .0837 units
moment of inertia of rod = ml² / 3 , m is mass of rod and l is length
= 2.95 x .95² / 3
I₁ = .8874 units
moment of inertia of rod + putty
I₁ + mr²
m is mass of putty and r is distance where it sticks
I₂ = .8874 + .045 x (2 x .95 / 3)²
I₂ = .905
Applying conservation of angular momentum
angular momentum of putty = final angular momentum of rod+ putty
.0837 = .905 ω
ω is final angular velocity of rod + putty
ω = .092 rad /s .
Hello There!
From what i know, gravitational force increases if the mass is increased.
If the mass is being decreased, then i assume it will be B. Decreases.
Hope This Helps You!
Good Luck :)
- Hannah ❤
Can i get brainliest . i think it was used as a way of communication.
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
