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:
53.13 °
Explanation:
In order to do this, we just need to apply the following:
tanα = Dy/Dx
Where:
Vy: speed of the ball in the y axis.
Vx: speed of the ball in the x axis.
At this point we do not need the speed of the first ball after the collision because in that moment is already heading in the direction that we are looking for. Therefore, we just need to use the innitial data to calculate the direction which the first ball will go.
According to this, then:
tanα = (40/30)
tanα = 1.3333
α = tan⁻¹(1.3333)
<h2>
α = 53.13°</h2>
This means that the final direction of the first ball is 53.13° and in the x axis because the starting momentum of this ball in the x axis has not dissapeared.
Hope this helps
Earth's atmosphere blocks many types of light including gamma, x-rays most ultraviolet and infrared. So optical telescopes that use visible light and ultraviolet telescopes that are used to study very hot stars are much less effective on Earth.
Answer:
Radiation
Explanation:
The sun energy reaches us by Radiation.
Answer:
1.04μT
Explanation:
Due to both wires have opposite currents, the magnitude of the total magnetic field is given by
I: electric current = 10A
mu_o: magnetic permeability of vacuum = 4pi*10^{-7} N/A^2
r1: distance from wire 1 to the point in which B is measured.
r2: distance from wire 2.
The distance between wires is 40cm = 0.4m. Hence, r1=0.2m r2=0.6m
By replacing in the formula you obtain:
hence, the magnitude of the magnetic field is 1.04μT
