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?
Complete Question
You are performing a double slit experiment very similar to the one from DL by shining a laser on two nattow slits spaced 
meters apart. However, by placing a piece of crystal in one of the slits, you are able to make it so that the rays of light that travel through the two slits are Ï out of phase with each other (that is to say, Ao,- ). If you observe that on a screen placed 4 meters from the two slits that the distance between the bright spot closest to center of the pattern is 1.5 cm, what is the wavelength of the laser?
Answer:
The wavelength is 
Explanation:
From the question we are told that
The distance of slit separation is 
The distance of the screen is 
The distance between the bright spot closest to the center of the interference is 
Generally the width of the central maximum fringe produced is mathematically represented as
=> 
=> 
=>
Check this Light doesn't have mass or gravity right?
So if it doesn't have mass or gravity so light can only affect objects with mass
Does that make sense?
The black hole has gravity and remember light doesn't have gravity so does it affect the light?
To answer that yes, and since light doesn't have gravity it gets "pulled" into the black hole
I hope this helps you
Answer:it helps get everyone in the same page
Explanation:
This is important so everyone knows what they should be doing
The capacitance of a capacitor is the ratio of the stored charge to its potential difference, i.e.
C = Q/ΔV
C is the capacitance
Q is the stored charge
ΔV is the potential difference
Rearrange the equation:
ΔV = Q/C
We also know the capacitance of a parallel-plate capacitor is given by:
C = κε₀A/d
C is the capacitance
κ is the capacitor's dielectric constant
ε₀ is the electric constant
A is the area of the plates
d is the plate separation
If we substitute C:
ΔV = Qd/(κε₀A)
We assume the stored charge and the area of the plates don't change. Then if we double the plate spacing, i.e. we double the value of d, then the potential difference ΔV is also doubled.
