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2 years ago

A small object is attached to a horizontal spring and set in simple harmonic motion with amplitude A and period T .

Physics

1 answer:

kupik [55]2 years ago

4 0

Answer:

t = 3/2T

To find how long it takes to cover a total distance of 6A, we need to find the time it takes to cover a distance A then multiply by 6.

The step to the solution is given below in the attachment.

Explanation:

Thank you for reading

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The weight of an astronaut plus his space suit on the Moon is only 250 N. (a) How much does the suited astronaut weigh on Earth?

Evgesh-ka [11]

Answer:

a)1500N

b)153.06kg

Explanation:

F = ma

g(moon) = is the acceleration due to gravity on the moon

g(earth) is the acceleration due to gravity on the earth

g(moon) = 1/6g(earth)

g(earth) =6g(moon)

F(gearth) = mg(earth)

= m 6g(moon)

= 6 × 250

= 1500N

b) F(gearth) = mg(earth)

m = F /g

= 1500/9.8

= 153.06kg

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3 years ago

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A rock, initially at rest with respect to Earth and located an infinite distance away is released and accelerates toward Earth.

lianna [129]

Answer:

the rock speed is increased

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3 years ago

3. A certain wire, 3 m long, stretches by 1.2 mm when under tension of 200 N. By how much does

nikitadnepr [17]

Answer:

The extension of the second wire is $e_2 = 0.0024 \ m = 2.4 mm$

Explanation:

From the question we are told that

The length of the wire is $L = 3 \ m$

The elongation of the wire is $e = 1.2mm = \frac{1.2}{1000} = 0.0012 m$

The tension is $F = 200 \ N$

The length of the second wire is $L_2 = 6 \ m$

Generally the Young's modulus(Y) of this material is

$Y = \frac{stress}{strain }$

Where $stress = \frac{F}{A}$

Where A is the area which is evaluated as

$A = \pi r^2$

and $strain = \frac{extention}{length} = \frac{e}{L}$

$Y = \frac{\frac{F}{\pi r^2 } }{ \frac{e}{L} }$

Since the wire are of the same material Young's modulus(Y) is constant

So we have

$\frac{F * L }{r^2 e} = \pi * Y = constant$

$F * L = constant * r^2 e$

Now the ration between the first and the second wire is

$\frac{F_1}{F_2} * \frac{L_1}{L_2} = \frac{r*2_1}{r^2} * \frac{e_1}{e_2}$

Since tension , radius are constant

We have

$\frac{L_1}{L_2} = \frac{e_1}{e_2}$

substituting values

$\frac{3}{6} = \frac{0.0012}{e_2}$

$0.5 e_2 = 0.0012$

$e_2 = \frac{ 0.0012 }{0.5}$

$e_2 = 0.0024 \ m = 2.4 mm$

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3 years ago

What is the Unit used to measure voltage

muminat

The derived unit for voltage is named volt.

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Particles with more energy move ___ than particles with less energy?

Hitman42 [59]

Particle with more energy move SLOWER than particles with less energy.

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2 years ago

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