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3 years ago

How to draw the ray diagram? Pls answer

Physics

1 answer:

weqwewe [10]3 years ago

7 0

1. Pick a point on the top of the object and draw two incident rays traveling towards the mirror. Using a straight edge, accurately draw one ray so that it passes exactly through the focal point on the way to the mirror. Draw the second ray such that it travels exactly parallel to the principal axis.

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Vesna [10]

Complete Question

A football coach walks 18 meters westward, then 12 meters eastward, then 28 meters westward, and finally 14 meters eastward.

From this motion what is the distance covered

What is the magnitude and direction of the displacement

Answer:

$D = 72 \ m$

Magnitude

$d = - 20 \ m$

Direction

West

Explanation:

From the question we are told that

The first distance covered westward is $d_w_1 = 18 \ m /tex] The first distance covered eastward is [tex]d_e1 = 12 \ m /tex] The second distance covered westward is [tex]d_w_2 = 28 \ m /tex] The second distance covered eastward is [tex]d_e2 = 14 \ m /tex] Generally the distance covered is mathematically represented as [tex]D = d_w1 + d_w2 + d_e1 + d_e2$

=> $D = 18 + 28 + 12 + 14$

=> $D = 72 \ m$

For the second question eastward is in the direction of the positive x-axis so it would be positive and westward is in the direction of the negative x-axis so it would be negative

The magnitude of the displacement is

$d = -d_w1 -d_w2 + d_e1 + d_e2$

=> $d = -18-28 + 12 + 14$

=> $d = - 20 \ m$

The direction is west

7 0

3 years ago

If for every action force an equal and opposite reaction force exlsts, how can anything ever be accelerated?

klasskru [66]

Answer:

Explanation:

The net force is the force between action and reaction and when this forces are not the same an acceleration is spurred.

3 0

3 years ago

Describe what a hydrogen bond is. 6th grade answer

r-ruslan [8.4K]

Answer:

<em>Hydrogen bond is the attractive force between the hydrogen attached electronegative atom </em>

Explanation:

8 0

3 years ago

a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?

julsineya [31]

Answer:

Approximately $7.0\; \rm m \cdot s^{-1}$ .

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant $g = 9.81 \; \rm m \cdot s^{-2}$ near the surface of the earth.

For an object that is accelerating constantly,

$v^2 - u^2 = 2\, a \, x$ ,

where

$u$ is the initial velocity of the object,
$v$ is the final velocity of the object.
$a$ is its acceleration, and
$x$ is its displacement.

In this case, $x$ is the same as the change in the ball's height: $x = 2.5\; \rm m$ . By assumption, this ball was dropped with no initial velocity. As a result, $u = 0$ . Since the ball is accelerating due to gravity, $a = 9.81\; \rm m \cdot s^{-2}$ .