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2 years ago

11

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

1 answer:

DiKsa [7]2 years ago

8 0

The correct answer is 5.00 m/s^2

Explanation:
According to Newton's second law:
F = ma ----- (1)

Where F = net force acting on the body = 125N
m = mass of the body = 25N
a = Acceleration of the body due to the applied force = ?

Plug-in the values in equation (1):
(1) => a = F/m
a = 125N/25kg
Hence a = 5.00 m/s^2

-i

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3 years ago

A 1500kg car traveling at 25m/s skids to a stop. The force of friction between the tires and the road is 10500N. How far does th

ale4655 [162]

44.64m

Explanation:

Given parameters:

Mass of the car = 1500kg

Initial velocity = 25m/s

Frictional force = 10500N

Unknown:

Distance moved by the car after brake is applied = ?

Solution:

The frictional force is a force that opposes motion of a body.

To solve this problem, we need to find the acceleration of the car. After this, we apply the appropriate motion equation to solve the problem.

-Frictional force = m x a

the negative sign is because the frictional force is in the opposite direction

m is the mass of the car

a is the acceleration of the car

a = $\frac{frictional force}{mass}$ = $\frac{10500}{1500}$ = -7m/s²

Now using;

V² = U² + 2as

V is the final velocity

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a is the acceleration

s is the distance moved

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learn more:

Velocity problems brainly.com/question/10932946

#learnwithBrainly

7 0

3 years ago

Compare the magnitude of the magnetic field at the center of a circular current loop of radius 30 mm with the magnitude of the m

NISA [10]

Answer:

Ratio of magnetic field will be $\frac{B}{B'}=0.0055$

Explanation:

We have given radius of the loop r = 30 mm = 0.03 m

We know that magnetic field at the center of the loop is given by

$B=\frac{\mu _0i}{2r}$ ---------eqn 1

Number of turns in the solenoid is given as n = 3 turn per mm = 3000 turn per meter

We know that magnetic field due to solenoid is given by

$B'=n\mu _0i$ -------------eqn 2

Now dividing eqn 1 by eqn 2

$\frac{B}{B'}=\frac{\frac{\mu _0i}{2r}}{\mu _0ni}=\frac{1}{2nr}=\frac{1}{2\times 3000\times 0.03}=0.0055$

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