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3 years ago

A body accelerate uniformly from rest at 0.2m/s for one-fifth of a minute. Calculate the distance covered by the body.

Physics

1 answer:

PtichkaEL [24]3 years ago

4 0

Answer:

The distance covered by the body is 14.4 m.

Solution,

Initial velocity(u)= 0 m/s( according to Question, the body starts from rest.that's why initial velocity(u) becomes zero.

Acceleration (a)= 0.2 m/s

Time (t):

$\frac{1}{5} minute = \frac{1}{5 } \times 60 \: seconds$

$= 12 \: seconds$

Now,

Applying third equations of motion:

$s = ut + \frac{1}{2} a {t}^{2} \\ s = 0 \times 12 + \frac{1}{2} \times 0.2 \times {(12)}^{2} \\s = 0 + 14.4 \\ s = 14.4 \: metre$

Thus, the distance covered by the body is 14.4 metre.

Further more information:

<h3>Application of equation of motion in different situation:</h3>

When a certain object comes in motion from rest, in the case, initial velocity(u)= 0 m/s
When a moving object comes in rest,in the case, final velocity(v)= 0 m/s
If the object is moving with uniform velocity, in the case, (u=v)
If any object is thrown vertically upward, in the case, acceleration (a)= -g
When an object is falling from certain height, in the case, u= 0 m/s
When an object is thrown vertically upwards in the case, final velocity at maximum height becomes 0.

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A uniformly charged ball of radius a and charge –Q is at the center of a hollowmetal shell with inner radius b and outer radius

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Answer:

r < a:

$E = \frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}$

r = a:

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

a < r < b:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

r = b:

$E = \frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$

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E = 0

r = c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

r < c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

Explanation:

Gauss' Law will be applied to each region to find the E-field.

$\int \vec{E}d\vec{a} = \frac{Q_{encl}}{\epsilon_0}$

An imaginary sphere is drawn with radius r, which is equal to the point where the E-field is asked. The area of this imaginary sphere is multiplied by E, and this is equal to the charge enclosed by this imaginary surface divided by ε0.

r<a:

Since the ball is uniformly charged and not hollow, then the enclosed charge can be found by the following method: If the total ball has a charge -Q and volume V, then the enclosed part of the ball has a charge Q_enc and volume V_enc. Then;

$\frac{Q}{V} = \frac{Q_{encl}}{V_{encl}}\\\frac{Q}{\frac{4}{3}\pi a^3} = \frac{Q_{encl}}{\frac{4}{3}\pi r^3}\\Q_{encl} = \frac{Qr^3}{a^3}$

Applying Gauss' Law:

$E4\pi r^2 = \frac{-Qr^3}{\epsilon_0 a^3}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}\\E = \frac{r}{4\pi a^3}\frac{Q}{\epsilon_0}$

The minus sign determines the direction of the field, which is towards the center.

At r = a:

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

At a < r < b:

The imaginary surface is drawn between the inner surface of the metal sphere and the smaller ball. In this case the enclosed charge is equal to the total charge of the ball, -Q.

$E4\pi r^2 = \frac{-Q}{\epsilon_0}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

At r = b:

$E = -\frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$

Again, the minus sign indicates the direction of the field towards the center.

At b < r < c:

The hollow metal sphere has a net charge of +2Q. Since the sphere is a conductor, all of its charges are distributed across its surface. No charge is present within the sphere. The smaller ball has a net charge of -Q, so the inner surface of the metal sphere must possess a net charge of +Q. Since the net charge of the metal sphere is +2Q, then the outer surface of the metal should possess +Q.

Now, the imaginary surface is drawn inside the metal sphere. The total enclosed charge in this region is zero, since the total charge of the inner surface (+Q) and the smaller ball (-Q) is zero. Therefore, the Electric region in this region is zero.

E = 0.

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The imaginary surface is drawn outside of the metal sphere. In this case, the enclosed charge is +Q (The metal (+2Q) plus the smaller ball (-Q)).

$E4\pi r^2 = \frac{Q}{\epsilon_0}\\E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

At r = c:

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This question is not complete.

The complete question is as follows:

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Explanation:

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1/T = number of revolutions per second

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A body accelerate uniformly from rest at 0.2m/s for one-fifth of a minute. Calculate the distance covered by the body.​

A body accelerate uniformly from rest at 0.2m/s for one-fifth of a minute. Calculate the distance covered by the body.