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:
10.85 m/s
Explanation:
m = 1200 kg, h1 = 19 m, h2 = 13 m
Let v be the velocity
Use the conservation of energy
Potential energy at 19 m = Potential energy at 13 m + kinetic energy
m x g x h1 = m x g x h2 + 1/2 mv^2
m x g (h1 - h2) = 0.5 x m v^2
g (h1 - h2) = 0.5 v^2
9.8 (19 - 13) = 0.5 x v^2
v = 10.85 m/s
Work<span> done is generally referred in </span>relation<span> to the force applied while energy is used in reference to other factors such as heat. </span>Power<span> is defined as </span>work<span> done per unit time. Or energy per unit time. Eg: if you supply 50 joules over a period of 5 seconds, you have supplied 50/5=10 Watts of </span>power<span>.</span>
Answer:
0.14 J
Explanation:
The maximum velocity is the amplitude times the angular frequency.
vmax = Aω
ω = vmax / A
ω = (3.2 m/s) / (0.06 m)
ω = 53.3 rad/s
For a spring-mass system:
ω = √(k / m)
ω² = k / m
k = ω²m
k = (53.3 rad/s)² (0.050 kg)
k = 142 N/m
The elastic potential energy is:
EE = ½ kx²
EE = ½ (142 N/m) (0.044 m)²
EE = 0.14 J