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Reil [10]
3 years ago
5

A net force of 125 N accelerates a 25.0 kg mass. What is the resulting acceleration?

Physics
2 answers:
Neko [114]3 years ago
5 0

Answer: a=5 m/s^2

Explanation:

The acceleration of an object can be calculated by using Newton's second law:

F=ma

where

F is the net force applied on the object

m is the mass of the object

a is its acceleration

In this problem, we have F=125 N and m=25.0 kg, so we can rearrange the equation to calculate the acceleration:

a=\frac{F}{m}=\frac{125 N}{25.0 kg}=5 m/s^2

tester [92]3 years ago
4 0
We Know, F = m*a
Here, m = 0.25 Kg
F = 125 N

Substitute it into the expression, 
125 = 25 * a
a = 125/25
a = 5

So, your final answer is 5

Hope this helps!
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The electric potential at a certain location from a point charge can be represented by V. What is the value of the electric pote
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<h2>Answer:</h2>

If you triple the charge, the electric potential is 3V.

<h2>Explanation:</h2>

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Physics Question: Determine the work that is being done by tension in pulling the box 193.0 cm along the table. Also determine t
lozanna [386]

1) The work done by tension is 12.0 J

2) The final speed of the box is 1.57 m/s

Explanation:

The work done by a force on an object is given by:

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and of the displacement

For the box in this problem, we have:

F = 6.2 N is the force applied (the tension in the string)

d = 193.0 cm = 1.93 m is the displacement

\theta=0 assuming that the string is horizontal

Substituting, we find the work done by the tension:

W=(6.2)(1.93)(cos 0)=12.0 J

2)

In order to determine the final speed of the box, we need to determine its acceleration first.

Beside the tension, acting forward, the other force acting horizontally on the box is the force of friction, whose magnitude is

F_f = \mu mg

where

\mu=0.16 is the coefficient of friction

m = 2.8 kg is the mass of the box

g=9.8 m/s^2 is the acceleration of gravity

Substituting,

F_f = (0.16)(2.8)(9.8)=4.4 N

Therefore, the net force on the box is

F=6.2  N - 4.4 N = 1.8 N

And the acceleration can be found by using Newton's second law:

F=ma

where

F = 1.8 N is the net force

m = 2.8 kg is the mass of the box

a is the acceleration

Solving for a,

a=\frac{F}{m}=\frac{1.8}{2.8}=0.64 m/s^2

Now we can finally find the final speed using the suvat equation:

v^2-u^2=2as

where

v is the final speed

u = 0 is the initial speed

a=0.64 m/s^2 is the acceleration

s = 1.93 m is the displacement

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(0.64)(1.93)}=1.57 m/s

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