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____ [38]
3 years ago
12

Frame S' passes frame S in the usual way (positive directions). An object also moves in the positive direction. Which is true? T

he object's speed relative to S is greater than its speed relative to S'. The object's speed relative to S can be greater than or less than its speed relative to S', depending on the actual values. The object's speed relative to S is equal to its speed relative to S'. The object's speed relative to S is less than its speed relative to S'.
Physics
1 answer:
lisov135 [29]3 years ago
3 0

Answer:

The answer is "The object's speed relative to S can be greater than or less than its speed relative to S', depending on the actual values."

Explanation:

The S' frame and the object are moving in a positive direction. The object is moving with respect to the S frame so the S frame the rest frame

take the velocity of the object with respect to the rest frame as v and the velocity of the S' frame with respect S frame as v2

relative velocity of the object to the S' frame would be

Vrel = v2- v

This means the Vrel of the object with respect to the S' frame is less than the Vrel  of the object with respect to the S frame

However is the S' velocity is greater than that of the object then the Vrel of the object with respect to the S' frame is greater than the Vrel of the object with respect to the S frame.

This would mean the second option is the answer, the relative speed of the object depends on the actual values.

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3 years ago
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A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Ne
Furkat [3]

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz

Therefore, the frequency of this mode of vibration is 138.87 Hz

7 0
3 years ago
Calculate the translational speed of a cylinder when it reaches the foot of an incline 7.05 mm high. Assume it starts from rest
mestny [16]

Height is 7.05 m and not 7.05 mm

Answer:

9.603 m/s

Explanation:

We are dealing with rotation, so velocity of centre of mass is given by;

v_cm = Rω

Since we are working with a solid cylinder, moment of inertia of the cylinder is; I = ½mR²

Since it is rolled from the top to the bottom, at the top it will have potential energy(mgh) while at the bottom it will have kinetic energy (rotational plus translational kinetic energy).

Using conservation of energy, we have:

P.E = K.E_t + K.E_r

Formula for rotational and kinetic energy here are;

K.E_t = ½mv²

K.E_r = ½Iω²

mgh = ½mv² + ½Iω²

Since we want to find translational speed(v), let's get rid of ω.

Earlier, we saw that v_cm = Rω

Thus; ω = v/R

Also, we know that I = ½mR².

Thus;

mgh = ½mv² + ½(½mR²)(v/R)²

This gives;

mgh = ½mv² + ¼mv²

Divide through by m to get;

gh = v²(½ + ¼)

gh = ¾v²

Making v the subject gives;

v = √(4gh/3)

v = √((4 × 9.81 × 7.05)/3)

v = 9.603 m/s

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Answer:

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Answer:

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