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?
The electromagnetic spectrum is the system of frequencies that show electromagnetic radiation, respective wavelengths, and photon energies. Some examples of frequencies found on the electromagnetic spectrum are radio waves, microwaves, infrared, optical, ultraviolet, X-rays, and gamma-rays.
The gravitational force between the two balls is 
Explanation:
The magnitude of the gravitational force between two objects is given by:
where
:
is the gravitational constant
m1, m2 are the masses of the two objects
r is the separation between the objects
For the balls in this problem, we have
r = 0.74 m
Substituting into the equation, we find the gravitational force between the two balls:
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Answer:
The current in the rods is 171.26 A.
Explanation:
Given that,
Length of rod = 0.85 m
Mass of rod = 0.073 kg
Distance 
The rods carry the same current in the same direction.
We need to calculate the current
I is the current through each of the wires then the force per unit length on each of them is
Using formula of force
Where, m = mass of rod
l = length of rod
Put the value into the formula
Hence, The current in the rods is 171.26 A.
