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3 years ago

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"Suppose you tie a rock to the end of a 0.96 m long string and spin it in a horizontal circle with a constant angular velocity o

f 20.25 rad/s. If the tension in the string is 970 N, what is the mass of the rock (ignore gravity)

1 answer:

Ber [7]3 years ago

8 0

Answer:

The mass of the rock is $m = 2.46406 \ kg$

Explanation:

From the question we are told that

The length of the string is $l = 0.96 \ m$

The angular velocity is $w = 20.25 \ rad/s$

The tension on the string is $T = 970 \ N$

Generally the centripetal force acting on the rock is mathematically evaluated as

$T = mlw^2$

making m the subject of the formula

$m = \frac{T}{w^2 * l}$

substituting values

$m = \frac{970}{(20.25^2) * (0.96)}$

$m = 2.46406 \ kg$

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A playground merry-go-round has a radius of 4.6 m and a moment of inertia of 200 kg-m2 and turns with negligible friction about

tankabanditka [31]

Answer:

8050 J

Explanation:

Given:

r = 4.6 m

I = 200 kg m²

F = 26.0 N

t = 15.0 s

First, find the angular acceleration.

∑τ = Iα

Fr = Iα

α = Fr / I

α = (26.0 N) (4.6 m) / (200 kg m²)

α = 0.598 rad/s²

Now you can find the final angular velocity, then use that to find the rotational energy:

ω = αt

ω = (0.598 rad/s²) (15.0 s)

ω = 8.97 rad/s

W = ½ I ω²

W = ½ (200 kg m²) (8.97 rad/s)²

W = 8050 J

Or you can find the angular displacement and find the work done that way:

θ = θ₀ + ω₀ t + ½ αt²

θ = ½ (0.598 rad/s²) (15.0 s)²

θ = 67.3 rad

W = τθ

W = Frθ

W = (26.0 N) (4.6 m) (67.3 rad)

W = 8050 J

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3 years ago

How can the momentum of an object be changed?

pickupchik [31]

C. You’re welcome

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3 0

3 years ago

Determine the distance between a newly discovered planet and its single moon if the orbital period of the moon is 1.2 Earth days

Vinil7 [7]

Answer:

The distance is $r = 55430496 \ m$

Explanation:

From the question we are told that

The period of the moon $T = 1.2 days = 1.2 * 24 * 3600 = 103680 \ s$

The mass of the planet is $m_p = 9.38*10^{24} kg$

Generally the period of the moon is mathematically represented as

$T = 2 * \pi * \sqrt{ \frac{r^3 }{ G * m_p } }$

Here G is the gravitational constant with value

$G = 6.67 *10^{-11} \ N \cdot m^2/kg^2$

=> $T = 2 * \pi * \sqrt{ \frac{r^3 }{ G * m_p } }$

=> $103680 = 2 * 3.142 * \sqrt{ \frac{r^3 }{ 6.67*10^{-11} * 9.38*10^{24} } }$

=> $272218492.31 = \frac{r^3}{ 6.67 *10^{-11} * 9.38*10^{24}}$

=> $r = \sqrt[3]{ 1.7031241*10^{23}}$ j

=> $r = 55430496 \ m$

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3 years ago

A weightless spring is stretched 10 cm by a suspended 1-kg block. If two such springs are used to suspend the block, one spring

leva [86]

Answer:

total stretch of the double-length spring will be 20 cm

Explanation:

given data

length x1 = 10 cm

mass = 1 kg

mass = double = 2 kg

to find out

the total stretch of the double-length spring will be

solution

we can say here spring constant is

k = mg ............1

k is spring constant and m is mass and g is acceleration due to gravity

so for in 1st case and 2nd case with 1 kg mass and 2 kg mass

kx1 = mg .........................2

and

kx2 = 2mg ........................3

x is length

so from equation 2 and 3

$\frac{kx1}{kx2}= \frac{1mg}{2mg}$

$\frac{x1}{x2} = \frac{1}{2}$

$\frac{10}{x2} = \frac{1}{2}$

x2 = 20

so total stretch of the double-length spring will be 20 cm

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3 years ago

Which velocity graph shows the car moving toward the right (away from the origin) at a steady (constant) velocity?

slega [8]

Explanation:

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$\text{Therefore, only option 2,5,8 matches this requirement.}$

$\text{Since the car is moving to the right, which is defined as the positive direction in this question, the velocity (}v\text{) is also positive.}$

$\text{Therefore, only option 5 is correct.}$

6 0

3 years ago