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2 years ago

6

How is motion affected by change in mass of an object and forces applied?

1 answer:

densk [106]2 years ago

6 0

Newtons second law says that the acceleration of an object (produced by a net force) is directly proportional to that magnitude of the net force. E.g. F = ma
where F is the net force of an object, m is mass and a is acceleration.
For example, if an object had a large mass, there would have to be more force in order to move it than if it was lighter.
In a linear motion, if you pushed two objects, one slightly larger than the other, with the same force, the acceleration of the smaller object would be bigger than the larger one. So the motion (change in position over time), of the larger object would be seen as lesser than the smaller one (in a situation where both forces are equal).

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An object that completes 20 vibrations in 10 seconds has a frequency of

nika2105 [10]

Answer:

<em> The object has frequency of 2 Hz and time period of 0.5 s.</em>

Explanation:

<em>Frequency</em> is defined as number of oscillation per second ie back and forth swings done in single second.

Here it is given that the object oscillates 20 times in 10 seconds.

So f = $\frac{20}{10}$ = 2Hz

The <em>time period</em> is defined as time taken by the object to complete one full oscillation.

T = $\frac{1}{f}$

T= $\frac{1}{2}$ =0.5 s

<em>Thus the object has frequency of 2 Hz and time period of 0.5 s.</em>

7 0

2 years ago

Which of the following is evidence that supports the big bang theory?

Stella [2.4K]

Answer:

D

Explanation:

6 0

2 years ago

Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el

rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

$q_1 = 21.0\mu C$

$q_1 = 47.0\mu C$

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

$E = \frac{kq}{x}^2$

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

$\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2$

solving for x we get

$\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}$

x = 0.200 m

b)electric field at half way mean x =0.25

$E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}$

E = 3.84*10^{-4} N/C

6 0

3 years ago

Read 2 more answers

A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur

Nonamiya [84]

Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

Kindly check the diagram in the attached image below.

5 0

2 years ago

What happens to your image as you walk away from a plane mirror on a wall? It gets smaller.

IRISSAK [1]

Explanation:

It stays the same size

........

6 0

3 years ago