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olga2289 [7]
2 years ago
5

A meteor is falling towards the earth. The mass and the radius of the earth is 6×10^24 kg and 6.4×10^3 km respectively. What is

the height of the meteor from the earth surface at which the acceleration due to gravity is 4m/s^2
Physics
1 answer:
joja [24]2 years ago
8 0

Answer:

3600km

Explanation:

gravitational constant is the gravitational attraction between any two things

G = gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²

g = acceleration due to gravity = 4 m/s²

M = mass of the earth = 6 × 10²⁴

r = radius of Earth + height of meteor

Since F = mg and F = GMm/r², mg = GMm/r²

g = GM/r²

r² = GM/g

r = √(GM/g)

r = √((6.67 × 10⁻¹¹ Nm²/kg²) × (6 × 10²⁴ kg))/(4 m/s²)) =

1 × 10⁷ m = 1 × 10⁴ km

To calculate the height of the meteor above the earth,

subtract the radius of the earth from the calculated radius.

height of meteor = 1 × 10⁴ km - 6 × 10³ km = 4 × 10³ km

quora Meave Gilchrist

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An open organ pipe of length 0.47328 m and another pipe closed at one end of length 0.702821 m are sounded together. What beat f
sineoko [7]

Answer:

fb = 240.35 Hz

Explanation:

In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.

Open tube:

f=\frac{v_s}{2L}         (1)

vs: speed of sound = 343m/s

L: length of the open tube = 0.47328m

You replace in the equation (1):

f=\frac{343m/s}{2(0.47228m)}=362.36Hz      

Closed tube:

f'=\frac{v_s}{4L'}

L': length of the closed tube = 0.702821m

f'=\frac{343m/s}{4(0.702821m)}=122.00Hz

Next, you use the following formula for the beat frequency:

f_b=|f-f'|=|362.36Hz-122.00Hz|=240.35Hz

The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz

7 0
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