A believe that’s called a reference point.
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?
Answer:
4452.5 J.
Explanation:
The diver have both kinetic and potential energy.
Ek = 1/2mv² ................. Equation 1
Where Ek = Kinetic Energy of the diver, m = mass of the diver, v = velocity of the diver.
Given: m = 65 kg, v = 6.4 m/s.
Substitute into equation 1
Ek = 1/2(65)(6.4²)
Ek = 1331.2 J.
Also,
Ep = mgh ............................ Equation 2
Where Ep = Potential energy of the diver when its above the water, h = height of the diver above the water, g = acceleration due to gravity.
Given: m = 65 kg, h = 4.9 m, g = 9.8 m/s²
Substitute into equation 2.
Ep = 65(4.9)(9.8)
Ep = 3121.3 J.
Note: When she hits the water, the potential energy is converted to kinetic energy.
E = Ek+Ep
Where E = Kinetic energy of the diver when she hits the water.
E = 1331.2+3121.3
E = 4452.5 J.
Answer:
Object 3 has greatest acceleration.
Explanation:
Objects Mass Force
1 10 kg 4 N
2 100 grams 20 N
3 10 grams 4 N
4 1 kg 20 N
Acceleration of object 1,
Acceleration of object 2,
Acceleration of object 3,
Acceleration of object 4,
It is clear that the acceleration of object 3 is 
and it is greatest of all. So, the correct option is (3).
Answer:
Where 
represent the force for each of the 5 cases 
presented on the figure attached.
Explanation:
For this case the figure attached shows the illustration for the problem
We have an inverse square law with distance for the force, so then the force of gravity between Earth and the spaceship is lower when the spaceship is far away from Earth.
Th formula is given by:
Where G is a constant 
represent the mass for the earth
represent the mass for the spaceship
represent the radius between the earth and the spaceship
For this reason when the distance between the Earth and the Spaceship increases the Force of gravity needs to decrease since are inversely proportional the force and the radius, and for the other case when the Earth and the spaceship are near then the radius decrease and the Force increase.
Based on this case we can create the following rank:
Where represent the force for each of the 5 cases presented on the figure attached.
