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12345 [234]
3 years ago
7

g Can a rigid body experience any ACCELERATION when the resultant force acting on that rigid body is zero? Explain.Can a rigid b

ody experience any ACCELERATION when the resultant force acting on that rigid body is zero? Explain.
Physics
1 answer:
Advocard [28]3 years ago
5 0

Answer:

<em>No, a rigid body cannot experience any acceleration when the resultant force acting on the body is zero.</em>

Explanation:

If the net force on a body is zero, then it means that all the forces acting on the body are balanced and cancel out one another. This sate of equilibrium can be static equilibrium (like that of a rigid body), or dynamic equilibrium (that of a car moving with constant velocity)

For a body under this type of equilibrium,

ΣF = 0   ...1

where ΣF is the resultant force (total effective force due to all the forces acting on the body)

For a body to accelerate, there must be a force acting on it. The acceleration of a body is proportional to the force applied, for a constant mass of the body. The relationship between the net force and mass is given as

ΣF = ma   ...2

where m is the mass of the body

a is the acceleration of the body

Substituting equation 2 into equation 1, we have

0 = ma

therefore,

a = 0

this means that<em> if the resultant force acting on a rigid body is zero, then there won't be any force available to produce acceleration on the body.</em>

<em></em>

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When all individual forces acting upon an object are balanced, it is the natural tendency of the object to
sleet_krkn [62]

Answer:

- maintain its state of motion

- Keep its velocity constant (either at zero or non-zero value)

8 0
3 years ago
What is the sound intensity level in decibels? Use the usual reference level of I0 = 1.0×10−12 W/m2.
jek_recluse [69]

Answer:

L = 130 decibels

Explanation:

The computation of the sound intensity level in decibels is shown below:

According to the question, data provided is as follows

I = sound intensity = 10 W/m^2

I0 = reference level = 1 \times 10-12 W/m^2

Now

Intensity level ( or Loudness)is

L = log10 \frac{I}{10}

L = log10 \frac{10}{1\times 10^{-12}}

L = log10 \times 1013

= 13 \times 1 ( log10(10) = 1)

Therefore  

L = 13 bel

And as we know that

1 bel = 10 decibels

So,

The  Sound intensity level is

L = 130 decibels

5 0
4 years ago
Identical objects, Object X and Object Y, are tied together by a string and placed at rest on an incline, as shown in the figure
goblinko [34]

The slope of the velocity time graph of an object moving with constant acceleration is constant

It will take approximately <u>3 seconds</u> for the center of mass of Object X to reach point J near the bottom of the incline

The reason why the above time value is correct is given as follows:

Known parameters:

Initial velocity of the objects, u = 0

The graph  in the question is a straight line graph with data points

(0, 0), (0.5, 1.0), (1.0, 2), (3.0, 6), and (3.5, 7)

Given that the slope of the velocity-time graph is constant, we have that the acceleration is constant and is given as follows;

a = \dfrac{\Delta v}{\Delta t } = \dfrac{v_2 - v_1}{t_2 - t_1}

Therefore;

a =  \dfrac{6 - 2}{3.0 - 1.0} = 2

The acceleration, a ≈ 2 m/s²

The distance from the center of mass of the Object X to the point J near the bottom = 9 m

The equation for distance travelled is given as follows;

s = u\cdot t + \dfrac{1}{2} \cdot a \cdot t^2

Which gives;

9 =  0\times t + \dfrac{1}{2} \times 2 \times t^2 = t^2

t = √9 = 3

The time it will take the center of mass of Object X to reach point J near the bottom of the incline is t = <u>3 seconds</u>

Learn more about motion under constant acceleration here:

brainly.com/question/16391598

6 0
3 years ago
What is Acceleration due to gravity in cm/minute 2 ?​
Xelga [282]

Answer:

f

Explanation:

:P

5 0
3 years ago
How is the direction of light changed when it travels from an optically denser medium to an optically rarer medium????? please a
ANTONII [103]

Answer:

The light bends away from the normal

Explanation:

We can solve the problem by using Snell's law:

n_1 sin \theta_1 = n_2 sin \theta_2

where:

n_1 is the index of refraction of the first medium

n_2 is the index of refraction of the second medium

\theta_1 is the angle of incidence (angle between the incoming ray and the normal to the interface)

\theta_2 is the angle of refraction (angle between the outcoming ray and the normal to the interface)

We can rearrange the equation as

sin \theta_2 = \frac{n_1}{n_2}sin \theta_1

In this problem, light travels from an optically denser medium to an optically rarer medium, so

n_1 > n_2

Therefore, the term \frac{n_1}{n_2} is greater than 1, so

sin \theta_2 > sin \theta_1\\\rightarrow \theta_2 > \theta_1

which means that the angle of refraction is greater than the angle of incidence, and so the light will bend away from the normal.

4 0
3 years ago
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