Answer:
Let lo be the length of the rod in the frame in which it is at rest and s' is the frame which is moving with a speed 0.8c in a direction making an angle 60° with x-axis. The components of lo along and perpendicular to the direction of motion are lo cos 60° and lo sin 60° respectively.
Now length of the rod along the direction of motion
= lo cos 60°_/1-(0.8) 2/c2
= lo/2×0.6
= 0.3 lo.
Length of the rod perpendicular to the direction of motion.
= lo sin 60°
=_/3/2 lo
Length of moving rod
l = [(0.3lo)2+{lo_/3/2} 2] 1/2
= 0.916 lo.
Percentage contraction
= lo-0.916lo/lo×100
= 8.4%.
Explanation:
<h2><u><em>
Brainliest?</em></u></h2>
Answer:
block is being held in place on a frictionless surface. The block has a mass of 3.9 kg and is held in place on an incline of angle = 32° by a horizontal force F, as shown in the figure below.
Explanation:
Answer:
The work done on the object by the force in the 5.60 s interval is 40.93 J.
Explanation:
Given that,
Force 
Mass of object = 2.00 kg
Initial position 
Final position 
Time = 4.00 sec
We need to calculate the work done on the object by the force in the 5.60 s interval.
Using formula of work done
Put the value into the formula
Hence, The work done on the object by the force in the 5.60 s interval is 40.93 J.
The restoring force of the spring cancels the weight of the mass, so by Newton's second law
∑ F = F[spring] - mg = 0 ⇒ F[spring] ≈ 45.1 N
where m = 4.60 kg and g = 9.80 m/s². Then the spring constant is k such that by Hooke's law,
F[spring] = k x
where x = 0.0231 m. Then the spring constant is
k = F[spring]/x ≈ 1950 N/m
