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poizon [28]
3 years ago
15

A 4900 n gear box with a mass of 500kg falls to the ground. What is it’s acceleration due to gravity?

Physics
1 answer:
Lena [83]3 years ago
8 0

Answer:

9.8 m/s^2, towards the ground

Explanation:

For an object in free fall, the only force acting on it is the force of gravity. We can write it as

F=mg

where

F is called weight of the object

m is the mass of the object

g is the acceleration due to gravity

We can re-arrange the equation as

g=\frac{F}{m}

For the box in the problem, we know:

F = 4900 N

m = 500 kg

Therefore, we can find the value of g:

g=\frac{4900}{500}=9.8 m/s^2

And the direction is the same as the force of gravity (downward).

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Where will the spacecraft be when the gravitational forces acting on it are equal?
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It would be not be able to move yet it would be in the air

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You have seen magnets sticking to the refrigerator door, or maybe in your science class room. They attract metal items, for exam
DENIUS [597]

Answer:

B) shrinks

Explanation:

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3 0
3 years ago
The amplitude of a standing sound wave in a long pipe closed at the left end is sketched below.
ollegr [7]

(1) The harmonic number for the mode of oscillation is 3.

(2) The pitch (frequency) of the sound is 579.55 Hz

(3) The level of the water inside the vertical pipe is 0.1 m.

<h3>The harmonic number</h3>

The harmonic number for the mode of oscillation illustrated for the closed pipe is 3.

<h3>Frequency of the wave</h3>

The pitch (frequency) of the sound is calculated from third harmonic formula;

f = 3v/4L

where;

  • v is speed of sound
  • L is length of the pipe

f = (3 x 340) / (4 x 0.44)

f = 579.55 Hz

<h3>level of the water</h3>

wave equation for first harmonic of a closed pipe is given as

f  = v/(4L)

251.1  = 340/(4L)

4L = 340/251.1

4L = 1.35

L = 1.35/4

L = 0.34 m

level of water = 0.44 m - 0.34 m = 0.1 m

Thus, the level of the water inside the vertical pipe is 0.1 m.

Learn more about harmonics of closed pipes here: brainly.com/question/27248821

#SPJ1

3 0
1 year ago
A typical running track is an oval with 74-mm-diameter half circles at each end. A runner going once around the track covers a d
lisabon 2012 [21]

The centripetal acceleration a is 4.32 \times 10^-4 m/s^2.

<u>Explanation:</u>

The speed is constant and computing the speed from the distance and time for one full lap.

Given, distance = 400 mm = 0.4 m,       Time = 100 s.

Computing the v = 0.4 m / 100 s

                         v = 4 \times 10^-3 m/s.

radius of the circular end r = 37 mm = 0.037 m.

            centripetal acceleration a = v^2 / r

                                                        = (4 \times 10^-3)^2 / 0.037

                                                    a = 4.32 \times 10^-4 m/s^2.

6 0
3 years ago
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