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Oksi-84 [34.3K]
3 years ago
10

A container of rocks is brought back from the Moon's surface where the acceleration due to gravity is 162 meters per second if t

he total mass of the container of rocks is 650 kilograms, what is its weight on
Earth's surface
Physics
1 answer:
spin [16.1K]3 years ago
6 0

Answer:

6318 N

Explanation:

From the question given above, the following data were obtained:

Acceleration due to gravity of the moon (gₘ) = 1.62 m/s²

Mass (m) of container = 650 kg

Weight (W) of container on the earth =.?

Next, we shall determine the acceleration due to gravity of the earth. This can be obtained as follow:

Acceleration due to gravity of the moon (gₘ) = 1.62 m/s²

Acceleration due to gravity of the earth (gₑ) =.?

gₘ = 1/6 × gₑ

1.62 = 1/6 × gₑ

1.62 = gₑ /6

Cross multiply

gₑ = 1.62 × 6

gₑ = 9.72 m/s²

Finally, we shall determine the weight of the container on the earth as follow:

Mass (m) of container = 650 kg

Acceleration due to gravity of the earth (gₑ) = 9.72 m/s²

Weight (W) of container on the earth =.?

W = m × gₑ

W = 650 × 9.72

W = 6318 N

Therefore, the weight of the container on earth is 6318 N

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The acceleration due to gravity on earth will decrease as which of the following occurs. The mass of the object decreases. The d
pentagon [3]

Answer:

The distance of the object to the center of the earth increases.

Explanation:

The acceleration due to gravity on Earth is given by:

g=\frac{GM}{r^2}

where

G is the gravitational constant

M is the Earth's mass

r is the distance of the object from the Earth's centre

We notice that:

- g does not depend on the mass of the object

- g is inversely proportional to r

This means that if the distance of the object from the Earth's centre increases, g decreases. So, the correct option is

The distance of the object to the center of the earth increases.

4 0
3 years ago
Find the first three harmonics of a string of linear mass density 2. 00 g/m and length 0. 600 m when the tension in it is 50. 0
gavmur [86]

The first three harmonics of the string are 131.8 Hz, 263.6 Hz and 395.4 Hz.

<h3>Velocity of the wave</h3>

The velocity of the wave is calculated as follows;

v = √T/μ

where;

  • T is tension
  • μ is mass per unit length = 2 g/m = 0.002 kg/m

v = √(50/0.002)

v = 158.1 m/s

<h3>First harmonic or fundamental frequency of the wave</h3>

f₀ = v/λ

where;

  • λ is the wavelength = 2L

f₀ = v/2L

f₀ = 158.1/(2 x 0.6)

f₀ = 131.8 Hz

<h3>Second harmonic of the wave</h3>

f₁ = 2f₀

f₁ = 2(131.8 Hz)

f₁ = 263.6 Hz

<h3>Third harmonic of the wave</h3>

f₂ = 3f₀

f₂ = 3(131.8 Hz)

f₂ = 395.4 Hz

Thus, the first three harmonics of the string are 131.8 Hz, 263.6 Hz and 395.4 Hz.

Learn more about harmonics here: brainly.com/question/4290297

#SPJ1

6 0
1 year ago
What is sex And new born baby why connected to women's veins?
san4es73 [151]

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3 0
3 years ago
What do odometer and speedometer read ?
maksim [4K]
Odometer: tells you the distance traveled by vehicle since it was new (or when last reset)

Speedometer: tells you the velocity of the vehicle at that moment.
6 0
3 years ago
A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r
aivan3 [116]

Answer:

Explanation:

Given that,

Initial angular velocity is 0

ωo=0rad/s

It has angular velocity of 11rev/sec

ωi=11rev/sec

1rev=2πrad

Then, wi=11rev/sec ×2πrad

wi=22πrad/sec

And after 30 revolution

θ=30revolution

θ=30×2πrad

θ=60πrad

Final angular velocity is

ωf=18rev/sec

ωf=18×2πrad/sec

ωf=36πrad/sec

a. Angular acceleration(α)

Then, angular acceleration is given as

wf²=wi²+2αθ

(36π)²=(22π)²+2α×60π

(36π)²-(22π)²=120πα

Then, 120πα = 8014.119

α=8014.119/120π

α=21.26 rad/s²

Let. convert to revolution /sec²

α=21.26/2π

α=3.38rev/sec

b. Time Taken to complete 30revolution

θ=60πrad

∆θ= ½(wf+wi)•t

60π=½(36π+22π)t

60π×2=58πt

Then, t=120π/58π

t=2.07seconds

c. Time to reach 11rev/sec

wf=wo+αt

22π=0+21.26t

22π=21.26t

Then, t=22π/21.26

t=3.251seconds

d. Number of revolution to get to 11rev/s

∆θ= ½(wf+wo)•t

∆θ= ½(0+11)•3.251

∆θ= ½(11)•3.251

∆θ= 17.88rev.

5 0
3 years ago
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