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IgorLugansk [536]
3 years ago
13

A 10 kg computer accelerates at a rate of 5 m/s2. How much force was applied to the computer?​

Physics
2 answers:
worty [1.4K]3 years ago
8 0

Answer:

<h2>50 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 10 × 5

We have the final answer as

<h3>50 N</h3>

Hope this helps you

AURORKA [14]3 years ago
6 0

Answer:

<h2><u>Given </u><u>:</u><u>-</u><u> </u></h2>
  • Mass = 10 kg
  • Acceleration = 5 m/s²
<h2><u>To </u><u>find</u></h2>

Force applied

<h2>SoluTion</h2>

As we know that Force applied is the product of mass and acceleration

\sf \: force \:  = 5 \times 10

\sf \: force \: =  50 \: n

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Answer:

350 ft/s²

Explanation:

First, convert mph to ft/s.

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Given:

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v = 0 ft/s

t = 0.24 s

Find: a

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a = (v − v₀) / t

a = (0 ft/s − 85.1 ft/s) / 0.24 s

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Rounded to two significant figures, the magnitude of the acceleration is 350 ft/s².

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A 4.0 kg ball is traveling at 3.0 m/s and strikes a wall. The ball bounces off the wall with a velocity of 4.0 m/s in the opposi
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Answer:

280 N

Explanation:

Applying Newton's third second law of motion,

F = m(v-u)/t................... Equation 1

Where F = Magnitude of the average force on the ball during contact, v = final velocity of the ball, u = initial velocity of the ball, t = time of contact of the ball and the wall.

Note: Let the direction of the initial velocity of the ball be positive

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Substitute into equation 1

F = 4(-4-3)/0.1

F = 4(-7)/0.1

F = -28/0.1

F = -280 N.

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3 years ago
How large a band of frequencies does each television broadcasting channel get ?
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Answer:

Since the waves must carry a great deal of visual as well as audio information, each channel requires a larger range of frequencies than simple radio transmission. TV channels utilize frequencies in the range of 54 to 88 MHz and 174 to 222 MHz. (The entire FM radio band lies between channels 88 MHz and 174 MHz.)

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A speedboat increases its speed from 14.5 m/s to 29.3 m/s in a distance of 172 m.
dalvyx [7]

Answer:

a. Acceleration, a = 1.88 m/s²

b. Time, t = 7.87 seconds.

Explanation:

Given the following data;

Initial velocity, U = 14.5m/s

Final velocity, V = 29.3m/s

Distance, S = 172m

a. To find the acceleration of the speedboat;

We would use the third equation of motion;

V² = U² + 2aS

Substituting into the formula

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858.49 = 210.25 + 344a

344a = 858.49 - 210.25

344a = 648.24

a = 648.24/344

Acceleration, a = 1.88 m/s²

b. To find the time;

We would use the first equation of motion;

V = U + at

29.3 = 14.5 + 1.88t

1.88t = 29.3 - 14.5

1.88t = 14.8

Time, t = 14.8/1.88

Time, t = 7.87 seconds.

6 0
3 years ago
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