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lakkis [162]
3 years ago
9

Please HELP ME If the motion of an object changes, what must be true about the forces acting on that object?

Physics
2 answers:
Leokris [45]3 years ago
5 0

Answer: Unbalanced forces cause an object to start moving, stop moving, or change direction. Unbalanced forces acting on an object will change the object's motion. Unbalanced forces can also stop a moving object.

Explanation:

atroni [7]3 years ago
4 0

Answer:

If there is a net force acting on an object, the object will have an acceleration and the object's velocity will change. ... Newton's second law states that for a particular force, the acceleration of an object is proportional to the net force and inversely proportional to the mass of the object.

Explanation:

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What are the three ways acceleration can occur
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Answer:

Change in velocity, change in direction, change in both velocity and direction

Explanation:

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3 years ago
Acceleration is defined as the rate of change for which of the following? Question 4 options: time position velocity displacemen
vovangra [49]

Acceleration can be defined as the rate of change in the velocity of an object. Option C is correct.

<h3>What is Acceleration?</h3>
  • It is defined as the rate of change in velocity.
  • It can also be defined as the rate of change in position in a particular direction.

a =  \dfrac {v-u}t\ \ \ \ \rmor}\\\\a = \dfrac {\Delta v }t

Where,

a - acceleration

\Delta v - change in velocity

t - time

Therefore, acceleration can be defined as the rate of change in the velocity of an object.

Learn more about Velocity:

brainly.com/question/2239252

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2 years ago
1. Approximately how many people watch the March Madness tournament?
Stolb23 [73]
40 million
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3 years ago
An object with a mass M = 250 g is at rest on a plane that makes an angle θ = 30 o above the horizontal. The coefficient of kine
liubo4ka [24]

Answer:

v = 79.2 m/s

Solution:

As per the question:

Mass of the object, m = 250 g = 0.250 kg

Angle, \theta = 30^{\circ}

Coefficient of kinetic friction, \mu_{k} = 0.100

Mass attached to the string, m = 0.200 kg

Distance, d = 30 cm = 0.03 m

Now,

The tension in the string is given by:

Mgsin\theta + \mu_{k}Mgcos\theta + Ma = T        (1)

Also

T = m(g + a)

Thus eqn (1) can be written as:

Mgsin\theta + \mu_{k}Mgcos\theta + Ma = m(g - a)

Mgsin\theta + \mu_{k}Mgcos\theta + Ma = mg + ma

mg - Mgsin\theta - \mu_{k}Mgcos\theta = (M - m)a

a = \frac{0.2\times 9.8 - 0.250\times 9.8\times sin30^{\circ} - 0.1\times 0.250\times 9.8\times cos30^{\circ}}{0.250 - 0.200}

a = 10.45\ m/s^{2}

Now, the speed is given by the third eqn of motion with initial velocity being zero:

v^{2} = u^{2} + 2ad

where

u = initial velocity = 0

Thus

v = \sqrt{2ad}

v = \sqrt{2\times 10.45\times 0.03} = 0.792\ m/s

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