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2 years ago

Which type of mirror produces images that are always upright and at the same distance from the mirror as the object is?

Physics

2 answers:

Nezavi [6.7K]2 years ago

8 0

<h3><u>Answer;</u></h3>

Flat mirror

<h3><u>Explanation;</u></h3>

<em><u>A flat mirror or a plane mirror forms images that are always at equal distance as the object from the mirror. The images formed by a pane mirror are always virtual and the size of the image is the same as that of the object.</u></em>
<em><u>Plane mirrors and convex mirrors only produces virtual images. Concave mirrors are the only mirrors that are capable of forming real images and occurs only when the object is located a distance greater than a focal length of the mirror.</u></em>

Alex73 [517]2 years ago

7 0

Well, that would be a plane (flat) mirror
<span>provided that </span>
<span>the mirror and the object are oriented parallel to each other</span>

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Gauges all very according the vehicle , model make and year and other factors. By turning the vehicle to the ___ position you ca

AleksandrR [38]

Answer:

By turning the vehicle "ON" position you can check to see if the gauges light works.

When we switch ON or turn a key to ON the engine, we can find all the gauges working or not.

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2 years ago

If a 100N weight sits in an express elevator what will be the weight when the elevator is traveling at a constant speed

yarga [219]

This is a "trick" question.

If the elevator is traveling at constant speed, it means it is at rest. This means anything inside the elevator traveling at constant speed, weights the same as in an elevator not moving -also at rest-.

So the 100N weight's weight doesn't change in an elevator traveling at constant speed.

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2 years ago

A 2 kg book is pushed from rest to a final velocity of 3 m/s. The book travels 2 m. How much force was the book pushed with

Studentka2010 [4]

Explanation:

first you have to find accelerarion, it is given that the initial velocity(u) is 3 m/s, distance travelled(s) be 2m finall it came to rest so final velocity be 0m/s

now using the 3rd law of motion

v^2=u^2+2as

0=9+2a2

a= -9/4m/s^2

now force=mass×accelration

=2kg×(-9/4)m/s^2

=4.5 N

4.5 newton force applied on the book!

✌️:)

6 0

2 years ago

Read 2 more answers

A force of 20 N acts upon a 5 kg block. What is the acceleration of the object.

horsena [70]

Answer:

4m/s/s

Explanation:

a=f/m

m=5kg

f=20N

20/5=4

(N=kg-m/s/s)

7 0

2 years ago

Read 2 more answers

A student sits on a rotating stool holding two 3.09-kg masses. When his arms are extended horizontally, the masses are 1.08 m fr

schepotkina [342]

Answer:

The New angular speed is $w_f = 2.034 rad/s$

The Kinetic energy before the masses are pulled in is $KE_i = 3.101 \ J$

The Kinetic energy after the masses are pulled in is $KE_f = 8.192 \ J$

Explanation:

From the we are told that masses are 1.08 m from the axis of rotation, this means that

The radius $r =1.08m$

The mass is $m = 3.09\ kg$

The angular speed $w = 0.770 \ rad/sec$

The moment of inertia of the system excluding the two mass $I = 3.25 \ kg \cdot m^2$

New radius $r_{new} = 0.34m$

Generally the conservation of angular momentum can be mathematical represented as

$w_f = [\frac{I_i}{I_f} ]w_i .....(1)$

Where $w_f$ is the final angular speed

$w_i$ is the initial angular speed

$I_i$ is the initial moment of inertia

$I_f$ is the final moment of inertia

Moment of inertia is mathematically represented as

$I = m r^2$

Where I is the moment of inertia

m is the mass

r is the radius

So the Initial moment of inertia is given as

$I_i = moment \ of \ inertia \ of\ the \ two \ mass \ + 3.25 \ kg \cdot m^2$

$I_i = 2m r^2 + 3.25$

The multiplication by is because we are considering two masses

$I_i = 2 [(3.09)(1.08)^2] +3.25 = 10.46 kg \cdot m^2$

So the final moment of inertia is given as

$I_f = 2[(3.09)(0.34)^2] +3.25 = 3.96 \ kg \cdot m^2$

Substituting these values into equation 1

$w_f = [\frac{10.46}{3.96} ] * 0.77 = 2.034 \ rad/sec$

Generally Kinetic energy is mathematically represented in term of moment of inertia as

$KE = \frac{1}{2} * I * w^2$

Now considering the kinetic energy before the masses are pulled in,

$KE_i = \frac{1}{2} * I_i * w^2_i$

The Moment of inertia would be $I_i = 10.46 \ Kg \cdot m^2$

The Angular speed would be $w_i = 0.77 \ rad/s$

Now substituting these value into the equation above

$KE_i = \frac{1}{2} * (10.46) * (0.770)^2 = 3.101 J$

Now considering the kinetic energy after the masses are pulled in,

$KE_f = \frac{1}{2} * I_f * w^2_f$

The Moment of inertia would be $I_f = 3.96 \ Kg \cdot m^2$

The Angular speed would be $w_f = 2.034 \ rad/s$

Now substituting these value into the equation above

$KE_f= \frac{1}{2} *(3.96)(2.034)^2$

$= 8.192J$

8 0

2 years ago