Answer:
The speed of the rock when it is at height h/4 is 
.
Explanation:
At maximum height the final velocity of the rock is equal to 0. Let u is the initial velocity of the rock. Using the conservation of energy to find it as :
.......(1)
We need to find the speed of the rock when it is at height h/4. Let v' is the speed. Using 3rd equation of motion as :
here a = -g and s = h/4
Using equation (1) :
So, the speed of the rock when it is at height h/4 is . Hence, this is the required solution.
Answer:
A: The frequency of the vibration is 1.3329 Hz
B: The total energy of the vibration is 18.39375 J
Explanation:
The force of the man his weight causes the raft to sink, and that causes the water to put a larger upward force on the raft. This extra force is a restoring force, because it is in the opposite direction of the force put on the raft by the man. Then when the man steps off, the restoring force pushes upward on the raft, and thus the raft – water system acts like a spring, with a spring constant found as follows:
k= F/x = ((75 kg) * (9.81 m/s²))/(5*10^-2 m) = 14715 N/m
The frequency of the vibration is determined by the spring constant (k) and the mass of the raft (210kg).
fn = 1/2π * √(k/m) = 1/2π * √(14715 / 210) = <u>1.3329 Hz</u>
<u>The frequency of the vibration is 1.3329 Hz</u>
<u />
<u>b) </u>
Since the gravitational potential energy can be ignored, the total energy will be :
Etot = 1/2 k* A² = 1/2 * (14715 )*(0.05)² = 18.39375 J
<u>The total energy of the vibration is 18.39375 J</u>
Answer:
Its Greater potential energy because the air is high up and that makes high energy power
Explanation:
Answer:
1.a) 1 kJ
1.b) 4 kJ
ratio 1:4
1.c) 4 times as before
2.a) 3.33 m/s2
Explanation:
1.a) bicycle's velocity =Displacement/time
=100/20 m/s
=5 m/s
bicycler's KE =1/2 *mass*(velocity)^2
=1/2*80*5^2
=1000 J = 1 kJ
1.b) bicycle's new velocity =200/20 m/s
=10 m/s
bicycler's new KE =1/2*80*10^2
=4000 J = 4 kJ
Ratio= KE 1 :KE new
= 1 :4
1.c) when bicycler's speed was doubled it increased the KE by 4 times (2^2). because In KE we consider the square of the speed , so the factor we increase the speed , the KE will get increased with the square value of it
ex : speed is triple the prior value , then the KE is as 3^2 times as before. that is 9 times
2.a) car acceleration = (20-0)/6 m/s2
= 3.33 m/s2
a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s .The amplitude of the subsequent oscillations 48.13 cm/s
a 1.25 kilogram block is fastened to a spring with a 17.0 newtons per meter spring constant. Given that K is equal to 14 Newtons per meter and mass equals 10.5 kg. The block is then struck with a hammer by a student while it is at rest, giving it a speedo of 46.0 cm for a brief period of time. The required energy provided by the hammer, which is half mv squared, is transformed into potential energy as a result of the succeeding oscillations. This is because we know that energy is still available for consultation. So access the amplitude here from here. He will therefore be equal to and by. Consequently, the Newton's spring constant is 14 and the value is 10.5. The velocity multiplied by 0.49
Speed at X equals 0.35 into amplitude, or vice versa. At this point, the spirit will equal half of K X 1 squared plus half. Due to the fact that this is the overall energy, square is equivalent to half of a K square or an angry square. amplitude is 13 and half case 14 x one is 0.35. calculate that is equal to initial velocities of 49 squares and masses of 10.5. This will be divided in half and start at about 10.5 into the 49-square-minus-14. 13.42 into the entire square in 20.35. dividing by 10.5 and taking the square as a result. 231 6.9 Six centimeters per square second. 10.5 into 49 sq. 14. 2 into a 13.42 square entire. then subtract 10.5 from the result to get the square. So that is 48.13cm/s.
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This is incomplete question Complete Question is:
a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s . what are The amplitude of the subsequent oscillations?
