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stepan [7]
3 years ago
5

The acceleration of gravity on the surface of planet X is 2.95 times the acceleration of gravity on the surface of Earth and the

radius of planet X is 3.5 times the radius of Earth. What is the mass of planet X?
Physics
1 answer:
Mice21 [21]3 years ago
3 0

Answer:

The mass of planet X is 36.23 times that if earth.                                  

Explanation:

Given that,

The acceleration of gravity on the surface of planet X is 2.95 times the acceleration of gravity on the surface of Earth and the radius of planet X is 3.5 times the radius of Earth

The formula of acceleration due to gravity is given by :

g=G\dfrac{M}{r^2}

G is universal gravitational constant

M is mass of Earth and r is radius of Earth

Let g' is the acceleration due to gravity on the surface of planet X. So,

g'=G\dfrac{M'}{r'^2}  .......(1)

M' is the mass of planet X and r' is the radius of planet X.

So,

2.95g=G\dfrac{M'}{(3.5r)^2}  .....(2)

Dividing equation (1) and (2) we get :

\dfrac{g}{2.95g}=\dfrac{M}{M'}\times \dfrac{(3.5r)^2}{r^2}\\\\\dfrac{1}{2.95}=\dfrac{M}{M'}\times 12.25\\\\\dfrac{M}{M'}=\dfrac{1}{2.95\times 12.25}\\\\\dfrac{M}{M'}=0.027\\\\M'=36.23\ M

So, the mass of planet X is 36.23 times that if earth.

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Answer:

0.6983 m/s

Explanation:

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L₀ = Initial length = 11 cm = 0.11 m

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x = stretch in the spring = L - L₀ = 0.27 - 0.11 = 0.16 m

m = mass of the mass attached = 0.021 kg

v = speed of the mass

Using conservation of energy

Kinetic energy of mass = Spring potential energy

(0.5) m v² = (0.5) k x²

m v² = k x²

(0.021) v² = (0.4) (0.16)²

v = 0.6983 m/s

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4 years ago
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Answer:

152.

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1.085m

Explanation:

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A piano wire with mass 3.00 g and length 80.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and ampli
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Answer:

(a) P_{avg} = 0.22 W

(b) P_{avg,2} = 0.056 W

Explanation:

given information:

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tension, F = 25 N

length, l = 80 cm = 0.8 m

frequency, f = 120 Hz

amplitude, A = 1.6 mm = 0.0016 m

(a) the average power carried by the wave, P_{avg}

P_{avg} = \frac{1}{2}(√μF)ω²A²

where,

ω = 2πf = 2π120 = 754s^{-1}

μ = \frac{m}{l}

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thus,

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(b) What happens to the average power if the wave amplitude is halved.

based on the equation above, we know that the average power is proportional to the square amplitude. therefore

\frac{P_{avg,1} }{P_{avg,2} } = \frac{A_{1} ^{2} }{A_{2} ^{2} } }

\frac{P_{avg,1} }{P_{avg,2} } = \frac{A_{1} ^{2} }{(0.5A_{1} )^{2} } }

P_{avg,2} = \frac{0.22}{4}

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v = √KE ÷ 1/2 M

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