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?
7) 6, i believe, (Cu) 1 atom+ (S) 1 atom+(0)4 atoms.
Explanation:
p=mv
p=5.6×75
p= 420
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The wavelength of the standing wave at fourth harmonic is; λ = 0.985 m and the frequency of the wave at the calculated wavelength is; f = 36.84 Hz
Given Conditions:
mass of string; m = 0.0133 kg
Force on the string; F = 8.89 N
Length of string; L = 1.97 m
1. To find the wavelength at the fourth normal node.
At the fourth harmonic, there will be 2 nodes.
Thus, the wavelength will be;
λ = L/2
λ = 1.97/2
λ = 0.985 m
2. To find the velocity of the wave from the formula;
v = √(F/(m/L)
Plugging in the relevant values gives;
v = √(8.89/(0.0133/1.97)
v = 36.2876 m/s
Now, formula for frequency here is;
f = v/λ
f = 36.2876/0.985
f = 36.84 Hz
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<span>- Objects within the focal length will create real images. --> false
In fact, objects within the focal length create virtual images, as it can be seen in the ray diagrams here:
https://upload.wikimedia.org/wikipedia/commons/4/47/Concavemirror_raydiagram_F.svg
- Concave mirrors converge distant parallel light rays on the focal point. --> TRUE: the parallel rays (with respect to the mirror's axis) are reflected back into the focal point of the mirror, as it can be seen also from the previous picture.
- Concave mirrors can only create real images. --> FALSE: as it can be seen from the first picture, when the object is between the focus and the mirror, its image is virtual.
Concave mirrors can create real and virtual images. --> TRUE: concave mirrors can create real and virtual images, depending on the position of the object.
- Objects far away from concave mirrors will appear enlarged. --> FALSE:
as it can be seen from the ray diagram, the size of the image is smaller than the size of the object. https://upload.wikimedia.org/wikipedia/commons/d/d2/Concavemirror_raydiagram_2F.svg
- Objects between the center of curvature and the focal point will create real images.--> TRUE: as it can be seen from the ray diagram (2F corresponds to the center of curvature), the image in this case is on the same side of the object, so it is real. </span>https://upload.wikimedia.org/wikipedia/commons/9/91/Concavemirror_raydiagram_2FE.svg
