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Andreas93 [3]
3 years ago
10

What effect does an unbalanced force have on an object?

Physics
2 answers:
Vladimir79 [104]3 years ago
7 0

<u>Answer:</u> This force will cause the motion in an object.

<u>Explanation:</u>

Unbalanced force is defined as the forces which are not equal in magnitude. When two forces with different magnitudes are applied on an object from opposite directions, the object will start moving in the direction of the net force.

These forces result in the motion of an object.

Balanced forces are the forces which are equal in magnitude. When two forces with equal magnitude are applied on an object from opposite directions, the object will not start moving.

These forces does not result in the motion of an object.

Hence, this force will cause the motion in an object.

Mars2501 [29]3 years ago
5 0
<span>it either changes the direction of the object or it stops the object in motion.</span>
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While dragging a crate a workman exerts a force of 628 N. Later, the mass of the crate is increased by a factor of 3.8. If the w
Delvig [45]
Force applied = F = 628 N 
<span>Acceleration = a m/s² </span>
<span>Newton's 2nd law of motion : F = Ma </span>
<span> a = F/M -------- (1) </span>
<span>New mass of the crate = M1 = 3.8M kg </span>
<span>New acceleration = a1 = F/M1 = F/(3.8 M) ----- (2) </span>
<span>a1/a = {F/(3.8M)}/(F/M) = 1/3.8 = 10/38 = 5/19 ------- Answer</span>
6 0
4 years ago
Read 2 more answers
A car of 1000 kg with good tires on a dry road can decelerate (slow down) at a steady rate of about 5.0 m/s2 when braking. If a
vredina [299]

(a) 4.0 s

The acceleration of the car is given by

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time interval

For this car, we have

v = 0 (the final speed is zero since the car comes to a stop)

u = 20 m/s is the initial velocity

a=-5.0 m/s^2 is the deceleration of the car

Solving the equation for t, we find the time needed to stop the car:

t=\frac{v-u}{a}=\frac{0-(20 m/s)}{-5.0 m/s^2}=4 s

(b) 40 m

The stopping distance of the car can be calculated by using the equation

v^2 - u^2 = 2ad

where

v = 0 is the final velocity

u = 20 m/s is the initial velocity

a = -5.0 m/s^2 is the acceleration of the car

d is the stopping distance

Solving the equation for d, we find

d=\frac{v^2-u^2}{2a}=\frac{0^2-(20 m/s)^2}{2(-5.0 m/s^2)}=40 m

(c) -5.0 m/s^2

The deceleration is given by the problem, and its value is -5.0 m/s^2.

(d) 5000 N

The net force applied on the car is given by

F=ma

where

m is the mass of the car

a is the magnitude of the acceleration

For this car, we have

m = 1000 kg is the mass

a=5.0 m/s^2 is the magnitude of the acceleration

Solving the formula, we find

F=(1000 kg)(5.0 m/s^2)=5000 N

(e) 2.0\cdot 10^5 J

The work done by the force applied by the car is

W=Fd

where

F is the force applied

d is the total distance covered

Here we have

F = 5000 N

d = 40 m (stopping distance)

So, the work done is

W=(5000 N)(40 m)=2.0\cdot 10^5 J

(f) The kinetic energy is converted into thermal energy

Explanation:

when the breaks are applied, the wheels stop rotating. The car slows down, as a result of the frictional forces between the brakes and the tires and between the tires and the road. Due to the presence of these frictional forces, the kinetic energy is converted into thermal energy/heat, until the kinetic energy of the car becomes zero (this occurs when the car comes to a stop, when v = 0).

8 0
4 years ago
What is the relationship among
lawyer [7]
The bigger the object the greater the gravitational pull, so the farther away the big object is its gravitational force begins to decrease. Refer to the picture for more explanation.

4 0
3 years ago
Explain when a falling object is in free fall.
Ksenya-84 [330]
There is no gravity
3 0
4 years ago
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A box with a mass of 18 kg is pushed across the floor. It has coefficient of friction of 0.39. Calculate the force of friction i
Taya2010 [7]

Answer:

68.8 N

Explanation:

From the question given above, the following data were obtained:

Mass (m) of box = 18 Kg

Coefficient of friction (μ) = 0.39

Force of friction (F) =?

Next, we shall determine the normal force of the box. This is illustrated below:

Mass (m) of object = 18 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Normal force (N) =?

N = mg

N = 18 × 9.8

N = 176.4 N

Finally, we shall determine the force of friction experienced by the object. This is illustrated below:

Coefficient of friction (μ) = 0.39

Normal force (N) = 176.4 N

Force of friction (F) =?

F = μN

F = 0.39 × 176.4

F = 68.796 ≈ 68.8 N

Thus, the box experience a frictional force of 68.8 N.

3 0
3 years ago
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