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3 years ago

M=-2 the imagine equals what?

Physics

1 answer:

WITCHER [35]3 years ago

8 0

Answer:

Image is twice as large as the object, and inverted

Explanation:

When an object is placed in front of a mirror, the mirror produces an image.

The magnification of the image is a number telling how much the size of the image is enlarged/diminished with respect to the object.

It is given by

$M=\frac{y'}{y}$

where

M is the magnification

y' is the size of the image

y is the size of the object

In this problem,

M = -2

This means that:

$y'=My\\y' = -2y$

So, we can conclude the following:

- The size of the image is twice the size of the real object

- The image is also inverted, because of the presence of the negative sign in the equation

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grin007 [14]

Answer: 17.68 s

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0$ is the height of the ball when it hits the ground

$y_{o}=70 m$ is the initial height of the ball

$V_{o}=82m/s$ is the initial velocity of the ball

$t$ is the time when the ball strikes the ground

$g=9.8m/s^{2}$ is the acceleration due to gravity

Having this clear, let's find $t$ from (1):

$0=70m+(82m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}$ (2)

Rewritting (2):

$-\frac{1}{2}(9.8m/s^{2})t^{2}+(82m/s)t+70m=0$ (3)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (4)

Where:

$a=-\frac{1}{2}(9.8m/s^{2}$

$b=82m/s$

$c=70m$

Substituting the known values:

$t=\frac{-82 \pm \sqrt{82^{2}-4(-\frac{1}{2}(9.8)(70)}}{2a}$ (5)

Solving (5) we find the positive result is:

$t=17.68 s$

7 0

3 years ago

Simplify -2/3 - 3/5 A) -1/15 B) -19/15 C) -3/5 D) -5/8

Leto [7]

I got B,when you subtract 3/5 from NEGATIVE 2/3 it creates a negative 19 over a positive 15.

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A spring with constant k = 78 N/m is at the base of a frictionless, 30.0°-inclined plane. A 0.50-kg block is pressed against the

olasank [31]

Explanation:

The given data is as follows.

Spring constant (k) = 78 N/m, $\theta = 30^{o}$

Mass of block (m) = 0.50 kg

According to the formula of energy conservation,

mgh sin $\theta = \frac{1}{2}kx^{2}$

h = $\frac{1}{2} \times \frac{kx^{2}}{mg Sin \theta}$

= $\frac{78 \times 0.04}{2 \times 0.5 \times 9.8 \times 0.5}$

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Thus, we can conclude that the distance traveled by the block is 0.64 m.

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3 years ago

saul85 [17]

Answer:

B

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