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xxMikexx [17]
3 years ago
5

An object is placed in front of a concave mirror, between the center of curvature of the mirror and its focal point, as shown in

the diagram below. Three light rays are traced, along with their corresponding reflected rays. Which statement below best describes the image formed?
A. The image is real.

B. The image is right-side-up.

C. The image is reduced.

D. No image is formed.

Physics
1 answer:
weeeeeb [17]3 years ago
6 0
<h2>Answer: The image is real</h2>

A concave mirror has a reflective surface that is curved inwards.  This type of mirrors reflect the light making it converge in a focal point therefore they are used to focus the light. This occurs because the light is reflected with different angles, since the normal to the surface varies from one point to another of the mirror.

Now, the formation of the image will depend on the following conditions:

-<u>If the object is at a distance greater than the focal distance</u>, a <u>real</u> and <u>inverted image</u> is formed that may be larger or smaller than the object.  

<u>-If the object is at a distance smaller than the focal distance</u>, a <u>virtual  upright image</u> is formed that is larger than the object.  

As we can see in the figure attached, <u>the object is at a distance greater than the focal distance</u>, this means<u> the image is real</u>. This can be seen in the point where the rays converge.

In addition, this is an special case, <u>the object is between the center of curvature of the mirror and its focal point</u>, which means the image produced is also <u>inverted and larger than the object.</u>

<u></u>

Therefore, the correct option is A.

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Snowcat [4.5K]

Answer:

Compound.

Explanation:

6 0
4 years ago
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A race car starts from rest and accelerates uniformly to 69 mph in 4.5 s.
Marizza181 [45]
For this question we should apply
a = v^2 - u^2 by t
a = 69 - 0 by 4.5
a = 69 by 4.5
a = 15.33
a = 6.85 m/s^2

If the answer in option is near to answer then , you can mark it as correct.
.:. The acceleration is 6.9 m/s^2
5 0
3 years ago
a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the
tatuchka [14]
The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N
the force of the wind F, acting horizontally, with intensity
F=12 N
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
T \cos \alpha -F=0
T \sin \alpha -W=
By dividing the second equation by the first one, we get
\tan \alpha =  \frac{W}{F}= \frac{24.5 N}{12 N}=2.04
From which we find
\alpha = 63.8 ^{\circ}
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N




7 0
4 years ago
An object is made of glass and has the shape of a cube 0.13 m on a side, according to an observer at rest relative to it. Howeve
Vlad1618 [11]

Answer:

v=0.9833\ c

Explanation:

The density changes means that the length in the direction of the motion is changed.

Therefore,

$\text{Density} = \frac{m}{lwh}$

Given :

Side,  b = h = 0.13 m

Mass, m = 3.3 kg

Density = 8100 kg/m^3

So,

$8100=\frac{3.3}{l \times 0.13 \times 0.13}$

$l=\frac{3.3}{8100 \times 0.13 \times 0.13}$

l = 0.024 m

Then for relativistic length contraction,

$l= l' \sqrt{1-\frac{v^2}{c^2}}$

$0.024= 0.13 \sqrt{1-\frac{v^2}{c^2}}$

$0.184=  \sqrt{1-\frac{v^2}{c^2}}$

$0.033=  1-\frac{v^2}{c^2}}$

$\frac{v^2}{c^2}= 0.967$

$\frac{v}{c}=0.9833$

v=0.9833\ c

Therefore, the speed of the observer relative to the cube is 0.9833 c (in the units of c).

3 0
3 years ago
A bus starts from rest.if the acceleration is 2m/s square, find
MrMuchimi

Answer:

The velocity after 2 seconds can be found through:

V = u +a*t

Where V is final velocity, u is initial velocity, a is acceleration and t is time.

V = 0 + 2* 2= 4 meters/second

The distance (s) can be found through:

V^2= u^2 +2*a* s

Where V is final velocity, u is initial velocity, a is acceleration.

4^2= 0^2 + 2 *2*s

16= 0 + 4s

s= 4 meters

Distance (s) can also be found through:

s= ut + 1/2 at^2

s= 0+ 1/2 *2*2^2= 1 *2*2

s= 4 meters

Explanation:

3 0
2 years ago
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