Answer:
Δω = -5.4 rad/s
αav = -3.6 rad/s²
Explanation:
<u>Given</u>:
Initial angular velocity = ωi = 2.70 rad/s
Final angular velocity = ωf = -2.70 rad/s (negative sign is
due to the movement in opposite direction)
Change in time period = Δt = 1.50 s
<u>Required</u>:
Change in angular velocity = Δω = ?
Average angular acceleration = αav = ?
<u>Solution</u>:
<u>Angular velocity (Δω):</u>
Δω = ωf - ωi
Δω = -2.70 - 2.70
Δω = -5.4 rad/s.
<u> Average angular acceleration (αav):</u>
αav = Δω/Δt
αav = -5.4/1.50
αav = -3.6 rad/s²
Since, the angular velocity is decreasing from 2.70 rad/s (in counter clockwise direction) to rest and then to -2.70 rad/s (in clockwise direction) so, the change in angular velocity is negative.
Answer:
(a)0.0675 J
(b)0.0675 J
(c)0.0675 J
(d)0.0675 J
(e)-0.0675 J
(f)0.459 m
Explanation:
15g = 0.015 kg
(a) Kinetic energy as it leaves the hand
(b) By the law of energy conservation, the work done by gravitational energy as it rises to its peak is the same as the kinetic energy as the ball leave the hand, which is 0.0675 J
(c) The change in potential energy would also be the same as 0.0675J in accordance with conservation law of energy.
(d) The gravitational energy at peak point would also be the same as 0.0675J
(e) In this case as the reference point is reversed, we would have to negate the original potential energy. So the potential energy as the ball leaves hand is -0.0675J
(f) Since at the maximum height the ball has potential energy of 0.0675J. This means:
mgh = 0.0675
0.015*9.81h = 0.0675
h = 0.459 m
The ball would reach a maximum height of 0.459 m
Answer:
Explanation:
Given that,
5J work is done by stretching a spring
e = 19cm = 0.19m
Assuming the spring is ideal, then we can apply Hooke's law
F = kx
To calculate k, we can apply the Workdone by a spring formula
W=∫F.dx
Since F=kx
W = ∫kx dx from x = 0 to x = 0.19
W = ½kx² from x = 0 to x = 0.19
W = ½k (0.19²-0²)
5 = ½k(0.0361-0)
5×2 = 0.0361k
Then, k = 10/0.0361
k = 277.008 N/m
The spring constant is 277.008N/m
Then, applying Hooke's law to find the applied force
F = kx
F = 277.008 × 0.19
F = 52.63 N
The applied force is 52.63N
To solve the problem it is necessary to apply the Malus Law. Malus's law indicates that the intensity of a linearly polarized beam of light, which passes through a perfect analyzer with a vertical optical axis is equivalent to:
Where,
indicates the intensity of the light before passing through the polarizer,
I is the resulting intensity, and
indicates the angle between the axis of the analyzer and the polarization axis of the incident light.
Since we have two objects the law would be,
Replacing the values,
Therefore the intesity of the light after it has passes through both polarizers is
Answer: The correct answer is option B.
Explanation:
Mass of the sled = 10 kg
Initial speed of the sled = 2 m/s
Kinetic energy of the sled =
Work done by the sled = 20 joules
The work done by the friction will be in opposite direction and equal to the magnitude of the work done of the sled that - 20 J.
Hence, correct answer is option B.