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lesantik [10]
3 years ago
10

What is the mass of an object that experiences an acceleration of 5m/s^2 when a force of 500N acts on it?

Physics
1 answer:
earnstyle [38]3 years ago
3 0
Data;
m (mass) = ?
a (acceleration) = 5 m/s²
F (force) = 500 N

Formula:
F = m * a

Solving:
F = m*a
500 = m*5
500 = 5m
5m = 500
m =  \frac{500}{5}
\boxed{\boxed{m = 100\:Kg}}

Answer:
<span>The mass of an object you are experimenting with is 100 kg</span>
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Answer:

Net Force = 10 N

acceleration: a=0.817 \frac{m}{s^2}

Explanation:

1) The net force on the box is the applied force (F = 13 N) minus the friction force (f = 3 N), Therefore, net force = 10 N

See attached free body diagram.

2) To find the box's acceleration, we need first to find the mass of the box (since they are giving us its weight in Newtons). To do such we use the equation for weight and solve for the box's mass:

weight=m*g\\120 N = m*9.8\frac{m}{s^2} \\\frac{120}{9.8} kg = m\\m=12.24 kg

Now we can find the box's acceleration by using the equation for the net force:

F=m*a\\10N = 12.24 kg * a\\\frac{10}{12.24} \frac{m}{s^2} =a\\a=0.817 \frac{m}{s^2}

5 0
3 years ago
The velocity of sound in air at STP (Standard Temperature and Pressure) is approximately 344 m/s. If the frequency of the sound
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Answer:

0.066m

Explanation:

use the formula

velocity= frequency × wave length

344m/s = 5200Hz× wavelength

wavelength = 344m/s/5200Hz

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What is the value of the normal force if the coefficient of kinetic friction is 0.22 and the kinetic frictional force is 40 newt
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Hope this helps you!

5 0
4 years ago
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A 15.0-kg block is dragged over a rough, horizontal surface by a 70.0-N force acting at 20.0° above the horizontal. The block is
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Answer:

Explanation:

from the question we are told that

Load L=15kg

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3 0
3 years ago
A student holds a bike wheel and starts it spinning with an initial angular speed of 9.0 rotations per second. The wheel is subj
zloy xaker [14]

Answer:

T = 0.00889 N*m

Explanation:

Given the initial speed

Vo = 9.0rev/s.

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V = 6.5 rev/s.

V = Vo + a*t

Solve to acceleration knowing the initial velocity and the velocity at 10 s

6.5 rev/s - 9 rev/s = a*10s

a = -0.25 rev/s^2.

Now the solve the time at stop time so V=0

V = Vo + a*t

0 = 9.0 - 0.25 rev/s *t,

t = 36 s The Stopping time.

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The torque can be find using the acceleration using the equation

T = I*a

I = 1/2*m*r^2

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T = 0.00889 N*m

7 0
3 years ago
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