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3 years ago

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:

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

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

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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Hoochie [10]

Answer:

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t = 5.909 s

Explanation:

Given:

- Initial velocity of A v_i,a = 30 ft /s

- Initial distance s_o = 0

- Length of the exchange zone s_f = 65 ft

- Time taken t = 2.5 s

Start out by A's velocity as he gets to the end of the exchange zone.

Part a

Runner A decelerates at a uniform rate, so you can use the equation:

s_f = s_o + v_i,a*t + 0.5a_a*t^2

65 = 0 + 30*2.5 + 0.5*a_a*2.5^2

a_a = -20 / 2.5^2

a _a= -3.2 ft/s^2

Runner B accelerates at v_f,a as final velocity at a uniform rate, so you can use the equation:

v_f,b^2 - v_i,b^2 = 2*a_b*s

a_b = (30 -3.2*2.5)^2 /2*65

a_b = 3.723 ft/s^2

Part b

When Runner B should begin running:

t = (v_f,b - v_i,b) / a_b

t = (30 - 3.2*2.5) / 3.723

t = 5.909 s

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3 years ago

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Nuetrik [128]

Answer: D

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3 years ago

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3 years ago

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lorasvet [3.4K]

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THWB WB WQI B BW W;QQ QY G VQ Q VQVY VAV G UQ U

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3 years ago

A 250 kg flatcar 25 m long is moving with a speed of 3.0 m/s along horizontal frictionless rails. A 61 kg worker starts walking

Anna11 [10]

Answer:

x=31.09m

Explanation:

p1=p2

The momentum of flatcar and the momentum of the worker so

The velocity of the worker is:

$m_{f}*v_{f}=m_{w}*v_{w}\\\\v_{f}=\frac{m_{f}*v_{f}}{m_{w}}\\v_{f}=\frac{61kg*3.0\frac{m}{s}}{250kg}\\v_{f}=0.732\frac{m}{s}$

The total motion has a total velocity and is

$Vt=v_{w}+v_{f}\\Vt=0.732\frac{m}{s}+3.0\frac{m}{s}\\Vt=3.732\frac{m}{s}$

The time the worker take walking is

$t=\frac{x}{v_{w}}\\t=\frac{25m}{3\frac{m}{s}}=8.33s$

Now the total time and the total velocity determinate the motion of tha flatcar how far has moved

$x=t*Vt\\x=8.33s*3.732\frac{m}{s} \\x=31.09m$

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3 years ago

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

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