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3 years ago

Which of the following represents an image that is located behind a mirror?

Physics

2 answers:

9966 [12]3 years ago

3 0

Answer:

i think its -di........

Marina CMI [18]3 years ago

3 0

Explanation:

There are some conventions while solving the problems based on mirrors. All parameters are taken like x-y coordinate system.

All the measurements are done from the optical centre of the mirror.

$d_i$ is the image distance

$d_o$ is the object distance

$-d_i$ is the image distance which is formed behind the mirror.

$-d_o$ is the object distance when an object is placed behind the mirror.

So, $-d_i$ shows the image that is located behind the mirror. Hence, this is the required solution.

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A.) Determine the work done by Zach on the bull.

tresset_1 [31]

Explanation:

The unit of work done is in Joules

Work done is a physical quantity that is defined as the force applied to move a body through a particular distance.

Work is only done when the force applied moves a body through a distance.

Work done = Force x distance

The maximum work is done when the force is parallel to the distance direction.

The minimum work is done when the force is at an angle of 90° to the distance direction.

So to solve this problem;

multiply the force applied by Zack and distance through which the bull was pulled.

8 0

3 years ago

A block of mass 0.500 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x (F

Marrrta [24]

Answer:

Part a)

$x = 0.4 m$

Part b)

$v_i = 11.7 m/s$

Part c)

Speed is more than the required speed so it will reach the top

Explanation:

Part a)

As we know that there is no frictional force while block is moving on horizontal plane

so we can use energy conservation on the block

$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$

$\frac{1}{2}0.500(12^2) = \frac{1}{2}(450)x^2$

$x = 0.4 m$

Part b)

If the track has average frictional force of 7 N then work done by friction while block slides up is given as

$W_f = -7( \pi R)$

$W_f = -7(\pi \times 1.00)$

$W_f = -22 J$

work done against gravity is given as

$W_g = - mg(2R)$

$W_g = -(0.500)(9.8)(2\times 1)$

$W_g = -9.8 J$

Now by work energy equation we have

$\frac{1}{2}mv_i^2 + W_f + W_g = \frac{1}{2}mv_f^2$

$\frac{1}{2}0.5(12^2) - 9.8 - 22 = \frac{1}{2}(0.5)v_f^2$

$v_f = 4.1 m/s$

Part c)

now minimum speed required at the top is such that the normal force must be zero

$mg = \frac{mv^2}{R}$

$v = \sqrt{Rg}$

$v = 3.13 m/s$

so here we got speed more than the required speed so it will reach the top

5 0

3 years ago

Binding energy is the energy needed to

densk [106]

Answer:

Answer C

Explanation:

4 0

2 years ago

Do you even juul bro?

blsea [12.9K]

Answer:

Explanation:

5 0

2 years ago

Read 2 more answers

The reaction is at dynamic equilibrium.

DiKsa [7]

Answer:

Nitrogen and hydrogen combine at the same rate that ammonia breaks down.

Explanation:

7 0

2 years ago