We divide the thin rectangular sheet in small parts of height b and length dr. All these sheets are parallel to b. The infinitesimal moment of inertia of one of these small parts is
where
Now we find the moment of inertia by integrating from
to
The moment of inertia is
(from (-a/2) to
(a/2))
Answer:
The answers to the questions have been solved in the attachment.
Explanation:
The answers to part a to e are all contained in the attachment. For answer part b, temperature and frequency were assumed to be fixed or constant. V² is directly proportional to T telling us that variation in T gives us a square in the frequency variation. This tells us why it is difficult when both frequencies are on this side of the black body.
Answer: nuclear fusion
Explanation: give me a brainliest
Answer:
375 m.
Explanation:
From the question,
Work done by the frictional force = Kinetic energy of the object
F×d = 1/2m(v²-u²)..................... Equation 1
Where F = Force of friction, d = distance it slide before coming to rest, m = mass of the object, u = initial speed of the object, v = final speed of the object.
Make d the subject of the equation.
d = 1/2m(v²-u²)/F.................. Equation 2
Given: m = 60.0 kg, v = 0 m/s(coming to rest), u = 25 m/s, F = -50 N.
Note: If is negative because it tends to oppose the motion of the object.
Substitute into equation 2
d = 1/2(60)(0²-25²)/-50
d = 30(-625)/-50
d = -18750/-50
d = 375 m.
Hence the it will slide before coming to rest = 375 m
Range of a projectile motion is given by
R = v cos θ / g (v sin θ + sqrt(v^2 sin^2 θ + 2gy_0)); where R = 188m, θ = 41°, g = 9.8m/s^2, y_0 = 0.9
188 = v cos 41° / 9.8 (v sin 41° + sqrt(v^2 sin^2 41° + 2 x 9.8 x 0.9)) = 0.07701(0.6561v + sqrt(0.4304 v^2 + 17.64)) = 0.05053v + 0.07701sqrt(0.4304v^2 + 17.64)
0.07701sqrt(0.4304v^2 + 17.64) = 188 - 0.05053v
0.005931(0.4304v^2 + 17.64) = 35344 - 19v + 0.002553v^2
0.002553v^2 + 0.1046 = 35344 - 19v + 0.002553v^2
19v = 35344 - 0.1046 = 35343.8954
v = 35343.8954/19 = 1860 m/s
