Answer:
I think its B
Explanation:
Can u PLZZZ give me brainliest
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:
62 N
Explanation:
Sum of the forces on the toolbox:
∑F = ma
T − mg = ma
T = mg + ma
T = m (g + a)
T = (5.0 kg) (9.8 m/s² + 2.5 m/s²)
T = 61.5 N
Rounded to two significant figures, the force exerted by the rope is 62 N.
Answer:
576 joules
Explanation:
From the question we are given the following:
weight = 810 N
radius (r) = 1.6 m
horizontal force (F) = 55 N
time (t) = 4 s
acceleration due to gravity (g) = 9.8 m/s^{2}
K.E = 0.5 x MI x ω^{2}
where MI is the moment of inertia and ω is the angular velocity
MI = 0.5 x m x r^2
mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg
MI = 0.5 x 82.65 x 1.6^{2}
MI = 105.8 kg.m^{2}
angular velocity (ω) = a x t
angular acceleration (a) = torque ÷ MI
where torque = F x r = 55 x 1.6 = 88 N.m
a= 88 ÷ 105.8 = 0.83 rad /s^{2}
therefore
angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s
K.E = 0.5 x MI x ω^{2}
K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules
<span>The protons and neutrons are much more massive than the electrons.
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The electrons are simply too small to really change the mass, so they aren't counted.
