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3 years ago

The object represented by this graph is moving

Physics

2 answers:

natali 33 [55]3 years ago

6 0

Answer: the third one, toward the origin at a constant velocity.

Explanation:

Vinil7 [7]3 years ago

4 0

The object represented by this graph is moving toward the origin at constant velocity.

Option 3.

<u>Explanation:</u>

In the figure, x-axis is representing increase in the time and y-axis is presenting increase in the distance from bottom to up. But the line in the graph which is plotted is decreasing from high distance to small distance with increase in time. So this indicates that as the time is increasing, the distance is decreasing.

And the object is moving toward the origin as the distance of the object motion is found to decrease with increase of time as per the graph. But the slope of the graph is found to be almost constant, this indicates that the velocity of the object is constant. Thus, the object represented by this graph is moving toward the origin at constant velocity.

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3 years ago

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A climber pulls herself 8 meters upwards with a force of 150 Newtons. If it takes her 16 seconds to cover the 8 meters, how much

mr Goodwill [35]

Answer:

P = 75 W

Explanation:

given,

Distance, L = 8 m

Force,F = 150 N

Time, t = 16 s

Work by the climber

Work done = Force x displacement

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W = 150 x 8

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3 years ago

Please help me with both of them

8_murik_8 [283]

Answer:

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Explanation:

4 0

3 years ago

A convex lens has a focal length of 16.5 cm. Where on the lens axis should an object be placed in order to get a virtual, enlarg

alexandr402 [8]

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

magnification, m = 1.90

Solution:

Magnification of lens, m = - $\frac{v}{u}$

where

u = object distance

v = image distance

Now,

1.90 = $\frac{v}{u}$

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

$\frac{1}{F} = \frac{1}{u}+ \frac{1}{v}$

$\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}$

$\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}$

$\frac{1}{16.5} = \frac{ 0.90}{1.90u}$

u = 7.8 cm

Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.

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3 years ago

Use the motion map to answer the question.

Lana71 [14]

Answer:

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3 years ago