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: The observing friend will the swimmer moving at a speed of 0.25 m/s.
Explanation:
- Let <em>S</em> be the speed of the swimmer, given as 1.25 m/s
- Let
be the speed of the river's current given as 1.00 m/s.
- Note that this speed is the magnitude of the velocity which is a vector quantity.
- The direction of the swimmer is upstream.
Hence the resultant velocity is given as, 
= S — S 0
= 1.25 — 1
= 0.25 m/s.
Therefore, the observing friend will see the swimmer moving at a speed of 0.25 m/s due to resistance produced by the current of the river.
Answer:
The speed of the sound wave on the string is 545.78 m/s.
Explanation:
Given;
mass per unit length of the string, μ = 4.7 x 10⁻³ kg/m
tension of the string, T = 1400 N
The speed of the sound wave on the string is given by;
where;
v is the speed of the sound wave on the string
Substitute the given values and solve for speed,v,
Therefore, the speed of the sound wave on the string is 545.78 m/s.
The Pacific plate collides with and is subducted under four other crustal plates. The area where this happens has lots of volcanic activity. This area is called the Ring of Fire.
This is a place under the Pacific Ocean and it has numerous
volcanic eruptions and earthquakes and so this region is called the Ring of
Fire. In this place there are about 450 active and dormant volcanoes and it is
almost 75 percent of the volcanoes present in our earth.
