The drawing shows an object attached to an ideal spring, which is hanging from the ceiling. The unstrained length of the spring
is indicated. For purposes of measuring the height h that determines the gravitational potential energy, the floor is taken as the position where h = 0 m. The equilibrium position at which the object hangs stationary is identified as position 2. The object is set into vertical simple harmonic motion between positions 1 and 3. Identify the positions where the kinetic energy KE, the elastic potential energy EPE, and the gravitational potential energy GPE each have their maximum values during an oscillation cycle.
1 answer:
Image is missing so I have attached it.
Also, the options are missing and it is;
A) KE is has a maximum value at position 3. EPE has a maximum value at position 2. GPE has a maximum value at position 1.
B) KE is has a maximum value at position 1. EPE has a maximum value at position 2. GPE has a maximum value at position 3.
C) KE is has a maximum value at position 2. EPE has a maximum value at position 3. GPE has a maximum value at position 1.
D) KE is has a maximum value at position 1. EPE has a maximum value at position 3. GPE has a maximum value at position 2.
E) KE is has a maximum value at position 2. EPE has a maximum value at position 1. GPE has a maximum value at position 3.
Answer:
Option C is the correct answer which says; KE is has a maximum value at position 2. EPE has a maximum value at position 3. GPE has a maximum value at position 1.
Explanation:
If an object vibrates about its mean position, under the influence of a restoring force, such that restoring force is directly proportional to the displacement from the mean position, the motion of the object is called simple harmonic motion. During Simple harmonic motion, the sum of Kinetic and potential energy remains constant.
Now, Looking at the diagram, Kinetic Energy (KE) is maximum at position 2.
Looking at the options, only C and E agree with this.
Thus, our answer is either option C or E.
However, Option E is not going to be right because it says that GPE is at a maximum at position 3, which is not true as the maximum GPE will occur at position 1.
Thus,
Option C fulfills that and therefore will be the correct answer.
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Answer:
Explanation:
a ) If the image of this object is viewed with the eyepiece adjusted for minimum eyestrain (image at the far point of the eye) , the image from object lens must have been formed at the focus of eye lens . So the objective image must have been formed at 19.5 - 2.75 = 16.75 cm from the object lens.
b ) Let the object distance be u
For object lens
v = 16.75 cm , f = .35 cm
1/v - 1/u = 1/f
1/16.75 - 1/u = 1/ .35
.0597 - 1/u = 2.857
1/u = - 2.7973
u = .3575 cm
c ) Angular magnification
=
v₀ and u₀ are image and object distance for object lens , D = 25 cm and f_e is focal length of eye lens
= (16.75 / .3575) x( 25 / 2.75)
= 46.85 x 9.09
= 426
I am attaching the missing figure
If we assume 100% efficiency, this means that the input power of the transformer must be equal to the output power
We divide output/input
240,000/12,000 = 20
This means that the transformer has a turns ratio of 1:20
<span>ns/np = 20/1 = 20
The voltage is amplified 20 times by the transformer, but this also means that the current decreases by a factor of 20, so the total power remains the same.</span>
If we assume 100% efficiency, this means that the input power of the transformer must be equal to the output power
We divide output/input
240,000/12,000 = 20
This means that the transformer has a turns ratio of 1:20
<span>ns/np = 20/1 = 20
The voltage is amplified 20 times by the transformer, but this also means that the current decreases by a factor of 20, so the total power remains the same.</span>