In a contest, two tractors pull two identical blocks of stone thesame distance over identical surfaces. However, block A is movi
1 answer:
Answer:
d) Both blocks have had equal losses of energy to friction
Explanation:
As it is mentioned in the question that two tractors pull two same stone blocks having the identical distance over the same surfaces
Moreover, the block A is twice as fast than block B at the time of crossing the finish line
So based on the above information, it contains the losses of identical friction
And we also know that
Friction energy loss is
It would be the same for both the blocks
hence, the option d is correct
You might be interested in
Answer:
The magnitude of the total momentum is 21.2 kg m/s and its direction is 39.5° from the x-axis.
Explanation:
Hi there!
The total momentum is calculated as the sum of the momenta of the pieces.
The momentum of each piece is calculated as follows:
p = m · v
Where:
p = momentum.
m = mass.
v = velocity.
The momentum is a vector. The 200 g-piece flies along the x-axis then, its momentum will be:
p = (m · v, 0)
p = (0.200 kg · 82.0 m/s, 0)
p = (16.4 kg m/s, 0)
The 300 g-piece flies along the y-axis. Its momentum vector will be:
p =(0, m · v)
p = (0, 0.300 kg · 45.0 m/s)
p = (0, 13.5 kg m/s)
The total momentum is the sum of each momentum:
Total momentum = (16.4 kg m/s, 0) + (0, 13.5 kg m/s)
Total momentum = (16.4 kg m/s + 0, 0 + 13.5 kg m/s)
Total momentum = (16.4 kg m/s, 13.5 kg m/s)
The magnitude of the total momentum is calculated as follows:
The direction of the momentum vector is calculated using trigonometry:
cos θ = px/p
Where px is the horizontal component of the total momentum and p is the magnitude of the total momentum.
cos θ = 16.4 kg m/s / 21.2 kg m/s
θ = 39.3 (39.5° if we do not round the magnitude of the total momentum)
Then, the magnitude of the total momentum is 21.2 kg m/s and its direction is 39.5° from the x-axis.
Answer:
0.5% per oscillation
Explanation:
The term 'damped oscillation' means an oscillation that fades away with time. For Example; a swinging pendulum.
Kinetic energy, KE= 1/2×mv^2-------------------------------------------------------------------------------------------------------------(1).
Where m= Mass, v= velocity.
Also, Elastic potential energy,PE=1/2×kX^2----------------------------------------------------------------------------------------------------------------------(2).
Where k= force constant, X= displacement.
Mechanical energy= potential energy (when a damped oscillator reaches maximum displacement).
Therefore, we use equation (3) to get the resonance frequency,
W^2= k/m--------------------------------------------------------------------------------------(3)
Slotting values into equation (3).
= 10/2.5.
= ✓4.
= 2 s^-1.
Recall that, F= -kX
F^2= (-0.1)^2
Potential energy,PE= 1/2 ×0.01
Potential energy= 0.05 ×100
= 0.5% per oscillation.
