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goblinko [34]
3 years ago
6

If the radius of a star increases by a factor of 6 the surface area of the star increases how many times?

Physics
1 answer:
Dennis_Churaev [7]3 years ago
8 0

Answer:

When radius of star increase by 6 factor then area of star will increase by a factor of 36.

Explanation:

As we know that

Area of star A given as

A=4\pi R^2

Where R is the radius of star.

Area of star when radius become 6 times

A'=4\pi (6R)^2

A'=4\pi \times 36\times R^2

So

A'=36A

We can say that when radius of star increase by 6 factor then area of star will increase by a factor of 36.

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In the diagram below, a 5 kg box is pushed from rest by an applied force of 15N for 22 seconds. What was the impulse caused by t
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Answer:

330Ns

Explanation:

Given parameters:

Mass of the box  = 5kg

Initial velocity  = 0m/s

Force applied  = 15N

Time  = 22s

Unknown:

Impulse  = ?

Final momentum  = ?

Solution:

The impulse of the box can be determined using the expression below:

     Impulse  = Force x time

    Impulse  = 15 x 22 = 330Ns

The final momentum is the same as this impulse

The impulse and the final momentum is 330Ns

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Two strings with linear densities of 5 g/m are stretched over pulleys, adjusted to have vibrating lengths of 0.50 m, and attache
HACTEHA [7]

Answer:

2.18 kg

Explanation:

The frequency of a wave in a stretched string f = n/2L√(T/μ) where n = harmonic number, L = length of string, T = tension = mg where m = mass of object on string and g = acceleration due to gravity = 9.8 m/s² and μ = linear density of string.

For string 1, its fundamental frequency f  is when n = 1. So,

f = 1/2L√(T/μ) =  1/2L√(mg/μ)

Now for string 1, L = 0.50 m, m = 20 kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f = 1/2L√(mg/μ)

f = 1/2 × 0.50 m√(20 kg × 9.8 m/s²/0.005 kg/m)

f = 1/1 m√(196 kgm/s²/0.005 kg/m)

f = 1/1 m√(39200 m²/s²)

f = 1/1 m × 197.99 m/s

f = 197.99 /s

f = 197.99 Hz

f ≅ 198 Hz

For string 2, at its third harmonic frequency f'  is when n = 3. So,

f' = 3/2L√(T/μ) =  3/2L√(mg/μ)

Now for string 2, L = 0.50 m, m = M kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f' = 3/2L√(Mg/μ)

f' = 3/2 × 0.50 m√(M × 9.8 m/s²/0.005 kg/m)

f' = 3/1 m√(M1960 m²/s²kg)

f' = 3/1 m√M√(1960 m²/s²kg)

f' = 3/1 m √M × 44.27 m/s√kg

f' = 132.81√M/s√kg

f' = 132.81√M Hz/√kg

Since the frequency of the beat heard is 2 Hz,

f - f' = 2 Hz

So, 198 Hz - 132.81√M Hz/√kg = 2 Hz

132.81√M Hz/√kg = 198 Hz - 2 Hz

132.81√M Hz/√kg = 196 Hz

√M Hz/√kg = 196 Hz/138.81 Hz

√M/√kg = 1.476

squaring both sides,

[√M/√kg] = (1.476)²

M/kg = 2.178

M = 2.178 kg

M ≅ 2.18 kg

8 0
3 years ago
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