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viva [34]
3 years ago
12

A body moving with an initial velocity of 30m/s accelerates uniformly at the rate of 10m/s . what is the distance covered during

the period​
Physics
1 answer:
nikdorinn [45]3 years ago
5 0

Answer:

The distance covered by the body is, S = 800 m

Explanation:

Given data,

The initial velocity of the body, u = 30 m/s

The acceleration of the body, a = 10 m/s²

Let the time period of travel be, t = 10 s

Using the II equations of motion,

                       S = ut + ½ at²

Substituting the given values,

                        S = 30 x 10 + ½ x 10 x 10²

                         S = 800 m

Hence, the distance covered by the body is, S = 800 m

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Which is true about a concave mirror? Incident rays that are parallel to the central axis are dispersed but will be perceived as
Reil [10]

Answer:

'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.

Explanation:

The question is incomplete, find the complete question in the comment section.

Concave mirrors is an example of a curved mirror. The outer surface of a concave mirror is always coated. On the concave mirror, we have what is called the central axis or principal axis which is a line cutting through the center of the mirror. The points located on this axis are the Pole, the principal focus and the centre of curvature. <em>The focus point is close to the curved  mirror than the centre of curvature.</em>

<em></em>

During the formation of images, one of the incident rays (rays striking the plane surface) coming from the object and parallel to the principal axis, converges at the focus point after reflection because all incident rays striking the surface are meant to reflect out. <em>All incident light striking the surface all converges at a point on the central axis known as the focus.</em>

Based on the explanation above, it can be concluded that 'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.

5 0
3 years ago
An object is moving along the x axis. At t = 5.6 s, the object is at x = +3.0 m and has a velocity of +5.7 m/s. At t = 8.5 s, it
Debora [2.8K]

The average acceleration between t = 5.6 s and t = 8.5 s is 2.31  m/s²

<h3>What is acceleration?</h3>

Acceleration is defined as the rate change of velocity with time.

acceleration a = (Δv) / (Δt)

An object is moving with initial velocity u =5.7 m/s and its final velocity v= -1.0 m/s.

Time taken for the change in speed, t= 8.5 - 5.6 = 2.9 seconds

The acceleration is given by

a = (-1 - 5.7)/ 2.9

a = -  2.31 m/s²

|a | = 2.31 m/s²

Thus, the object's acceleration is 2.31 m/s²

Learn more about acceleration.

brainly.com/question/12550364

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3 0
2 years ago
HURRY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
JulijaS [17]
Total amount of energy would remain constant according to law of conservation of energy. i.e., 50 Joules

In short, Your Answer would be Option C) <span>50 Joules because as energy converts from one form to another, it cannot be created or destroyed during the conversion.
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Hope this helps!

7 0
3 years ago
Read 2 more answers
In the diagram, a force of 20 newtons is applied to a block. The block is in dynamic equilibrium. What is the magnitude and dire
sladkih [1.3K]

Answer:B 20 newtons opposite to the direction of the applied force

Explanation:

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2 years ago
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A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate
Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is \bf{2.81 \times 10^{4}~n~C^{-1}}.

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and '\epsilon_{0}' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm

Part(c):

If we apply Gauss' law of electrostatics, then

&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}

3 0
3 years ago
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