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3 years ago

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A planet's moon travels in an approximately circular orbit of radius 7.0 ✕ 10⁷ m with a period of 6 h 38 min. Calculate the mass

of the planet from this information. ___ kg

1 answer:

natka813 [3]3 years ago

5 0

Answer:

3.56×10²⁶ Kg.

Explanation:

Note: The gravitational force is acting as the centripetal force.

Fg = Fc........................... Equation 1

Where Fg = gravitational Force, Fc = centripetal force.

Recall,

Fg = GMm/r²......................... Equation 2

Fc = mv²/r............................. Equation 3

Where M = mass of the planet, m = mass of the moon, r = radius of the orbit and G = Universal gravitational constant.

Substituting equation 2 and 3 into equation 1

GMm/r² = mv²/r

Simplifying the equation above,

M = v²r/G .............................. Equation 4.

The period of the moon in the orbit

T = 2πr/v

Making v the subject of the equation,

v = 2πr/T............................. Equation 5

where r = 7.0×10⁷ m, T = 6 h 38 min = (6×3600 + 38×60) s = (21600+2280) s

T = 23880 s, π = 3.14

v = (2×3.14×7.0×10⁷ )/23880

v = 18409 m/s

Also Given: G = 6.67×10⁻¹¹ Nm²/kg²

Also substituting into equation 4

M = 18409²×7.0×10⁷ /(6.67×10⁻¹¹)

M = 3.56×10²⁶ Kg.

Thus the mass of the planet = 3.56×10²⁶ Kg.

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Planets A and B have the same size, mass, and direction of travel, but planet A is traveling through space at half the speed of

Ganezh [65]

Answer:

B. You would weigh the same on both planets because their masses and the distance to their centers of gravity are the same.

Explanation:

Given that Planets A and B have the same size, mass.

Let the masses of the planets A and B are $m_A$ and $m_B$ respectively.

As masses are equal, so $m_A=m_B\cdots(i)$ .

Similarly, let the radii of the planets A and B are $r_A$ and $r_B$ respectively.

As radii are equal, so $r_A=r_B\cdots(ii)$ .

Let my mass is m.

As the weight of any object on the planet is equal to the gravitational force exerted by the planet on the object.

So, my weight on planet A, $w_A= \frac {Gm_Am}{r_A^2}$

my weight of planet B, $w_B=\frac {Gm_Bm}{r_B^2}$

By using equations (i) and (ii),

$w_B=\frac {Gm_Am}{r_A^2}=w_A$ .

So, the weight on both planets is the same because their masses and the distance to their centers of gravity are the same.

Hence, option (B) is correct.

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2 years ago

Planet Tatoone is about 1.7 AU from its Sun. Approximately how long will it take for light to travel from the Sun to Tatoone in

Radda [10]

Answer:

The value is $t = 14.129 \ minutes$

Explanation:

From the question we are told that

The distance of planet Tatoone is $d = 1.7 \ AU = 1.7 *1.496* 10^{11}=2.543*10^{11} \ m$

The speed of light is $c = 3.0*10^{8} \ m/ s$

Generally the time taken is mathematically represented as

$t = \frac{d}{c}$

=> $t = \frac{2.543*10^{11}}{3.0*10^{8} }$

=> $t = 847.7 \ s$

Now converting to minutes

$t = \frac{847.7}{60}$

=> $t = 14.129 \ minutes$

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3 years ago

As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th

NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

$v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek$

Conservation of Momentum:

$P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1$

Energy Balance:

$\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2} ....... Eq 2$

Substitute Eq 2 into Eq 1

$0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}$

Using Eq 1

$0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m$

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3 years ago

In order to determine the mass moment of inertia of a flywheel of radius 600 mm, a 12-kg block is attached to a wire that is wra

shtirl [24]

Answer:

Explanation:

Given that,

When Mass of block is 12kg

M = 12kg

Block falls 3m in 4.6 seconds

When the mass of block is 24kg

M = 24kg

Block falls 3m in 3.1 seconds

The radius of the wheel is 600mm

R = 600mm = 0.6m

We want to find the moment of inertia of the flywheel

Taking moment about point G.

Then,

Clockwise moment = Anticlockwise moment

ΣM_G = Σ(M_G)_eff

M•g•R - Mf = I•α + M•a•R

Relationship between angular acceleration and linear acceleration

a = αR

α = a / R

M•g•R - Mf = I•a / R + M•a•R

Case 1, when y = 3 t = 4.6s

M = 12kg

Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 4.6²

3 × 2 = 4.6²a

a = 6 / 4.6²

a = 0.284 m/s²

M•g•R - Mf = I•a / R + M•a•R

12 × 9.81 × 0.6 - Mf = I × 0.284/0.6 + 12 × 0.284 × 0.6

70.632 - Mf = 0.4726•I + 2.0448

Re arrange

0.4726•I + Mf = 70.632-2.0448

0.4726•I + Mf = 68.5832 equation 1

Second case

Case 2, when y = 3 t = 3.1s

M= 24kg

Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 3.1²

3 × 2 = 3.1²a

a = 6 / 3.1²

a = 0.6243 m/s²

M•g•R - Mf = I•a / R + M•a•R

24 × 9.81 × 0.6 - Mf = I × 0.6243/0.6 + 24 × 0.6243 × 0.6

141.264 - Mf = 1.0406•I + 8.99

Re arrange

1.0406•I + Mf = 141.264 - 8.99

1.0406•I + Mf = 132.274 equation 2

Solving equation 1 and 2 simultaneously

Subtract equation 1 from 2,

Then, we have

1.0406•I - 0.4726•I = 132.274 - 68.5832

0.568•I = 63.6908

I = 63.6908 / 0.568

I = 112.13 kgm²

8 0

3 years ago

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Answer:

<em>1</em><em>.</em><em>for </em><em>the </em><em>first </em><em>one </em><em>100c</em><em>e</em><em>n</em><em>t</em><em>i</em><em>m</em><em>e</em><em>t</em><em>e</em><em>r</em><em>s</em><em> </em><em>make </em><em>1</em><em> </em><em>meter </em><em>therefore</em>

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<em>=</em><em>3</em><em>.</em><em>4</em>

<em>the </em><em>answer </em><em>is </em><em>supposed</em><em> to</em><em> be</em><em> </em><em>3</em><em>.</em><em>4</em><em>,</em><em> maybe</em><em> </em><em>there's</em><em> </em><em>a </em><em>mistake</em><em> </em><em>with </em><em>the </em><em>question</em><em> </em><em>or </em><em>the </em><em>answer</em>

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