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3 years ago

A beam of light created by a concave mirror occurs when:

Physics

1 answer:

kozerog [31]3 years ago

4 0

The reflection law allows us to find that the correct statement is:

The angle of incidence is equal to the angle of reflection.

Geometric optics studies the geometric properties of light rays by surfaces. It has two laws that describe it:

The refraction law. This stable how a ray of light is refracted from one surface to another

The reflection law. Which states that when a ray is reflected off a surface the incident angle is equal to the reflected angle

$\theta_i = \theta_r$

Now we can analyze the expressions.

a) False. This only happens if the object is at infinity

b) True. This is the law of reflection that the angles are equal and it always complies

c) False the optical axis is a perpedicular line to the mirror, therefore the ray must be parallel to the mirror and is not reflected.

In conclusion using the reflection law we can find which statement is correct

The angle of incidence is equal to the angle of reflection.

Learn more here: brainly.com/question/15655610

You might be interested in

While unrealistic, we will examine the forces on a leg when one falls from a height by approximating the leg as a uniform cylind

Leno4ka [110]

Answer:

Part A: 7.75 m/s

Part B: 2330.8 kN

Part C: 24.03 kN

Part D: 4.8 kN

Part E: $1.7\times 10^{9} Dyn/cm^{2}$

Part F: Option D

Bending his legs increases the time over which the ground applies force, thus decreasing the force applied by the ground.

Explanation:

Part A

From the fundamental kinematic equation

$v^{2}=u^{2}+2gh$ where v is the velocity of the man just before hitting the ground, g is acceleration due to gravity, u is initial velocity, h is the height.

Since the initial velocity is zero hence

$v^{2}=2gh$

$v=\sqrt 2gh$

Substituting 10 m/s2 for g and 3 m for h we obtain

$v=\sqrt 2\times 10\times 3 =\sqrt 60= 7.745967\approx 7.75 m/s$

Part B 

Force exerted by the leg is given by

F=PA where P is pressure, F is force, A is the cross-section of the bone

$A=\frac {\pi d^{2}}{4}$

Substituting 2.3 cm which is equivalent to 0.023m for d and $1.7\times10^{8} N/m2$ for P we obtain the force as

$F=PA=1.7\times10^{8}*\frac {\pi (0.023)^{2}}{4}= 2330818.276\approx 2330.8 kN$

Part C

The fundamental kinematic equation is part (a) can also be written as

$v^{2}=u^{2}+2a\triangle x$ and making a the subject then

$a=\frac {v^{2}-u^{2}}{2\triangle x}$ where a is acceleration and $\triangle x$ is the change in length

Substituting the value obtained in part a, 7.75 m/s for v, u is zero and 1cm which is equivalent to 0.01 m for $\triangle x$ then

$a=\frac {7.75^{2}-0^{2}}{2\times 0.01}= 3003.125 m/s^{2}$

Force exerted on the man is given by

$F=ma=80\times 3003.125= 240250 N\approx 24.03 kN$

Part D

The fundamental kinematic equation is part (a) can also be written as

$v^{2}=u^{2}+2a\triangle h$ and making a the subject then

$a=\frac {v^{2}-u^{2}}{2\triangle h}$ where a is acceleration and $\triangle h$ is the change in height

Also, force exerted on the man is given by $F=ma=m\times \frac {v^{2}-u^{2}}{2\triangle h}$

Substituting 80 Kg for m, 50 cm which is equivalent to 0.5m for \triangle h and other values as used in part c

$F=ma=m\times \frac {v^{2}-u^{2}}{2\triangle h}=80\times \frac {7.75^{2}-0^{2}}{2\times 0.5}= 4805 N\approx 4.8 kN$

Part E

$P=1.7\times 10^{8}=1.7\times 10^{8}\times (\frac {10^{5} Dyn}{10^{4} cm^{2}}=1.7\times 10^{9} Dyn/cm^{2}$

Part F

Bending his legs increases the time over which the ground applies force, thus decreasing the force applied by the ground

7 0

3 years ago

A diagnostic sonogram produces a picture of internal organs by passing ultrasound through the tissue. In one application, it is

miss Akunina [59]

Answer:

1) $\lambda = 11\times 10^{-4} m$

2) $2.51\times 10^9 Pa$

Explanation:

Given data:

speed of sound v = 1540 m/s

frequency f = 1.40 MHz = 1.40 \times 10^6 Hz

density $\rho = 1060 kg/m^3$

1) we know that

$v = f\lambda$

$\lambda = \frac[v}{f} = \frac{1540}{1.40\times 10^6}$

$\lambda = 11\times 10^{-4} m$

2) we know that

$v= \sqrt{\frac{modulus}{density}}$

$v^2 = \frac{\gamma}{\rho}$

$\gamma = v^2 \rho$

$\gamma = 1540^2 \times 1060 = 2.51 \times 10^9 Pa$

6 0

4 years ago

An 82 g arrow is fired from a bow whose string exerts an average force of 95 n on the arrow over a distance of 77 cm. what is th

Zigmanuir [339]

You need to find speed, v.
And you (should) recognize the famous F=ma.

We've got data for F and m, so we can find a first!
$a = \frac{f}{m}$
a = 95N / 0.082 kg

Now from a we nees to find v. But how is v and a related?
$a = v \times \frac{dv}{dx}$
where dv is the change in speed (some unknown v minus zero since arrow is initially at rest). dx is the distance where the arrow experiences a force (77 cm!).

Now you know values for a, dv, and dx... you can plug those in and find v!

5 0

3 years ago

Calculate the average travel time for each distance, and then use the results to calculate.

Eddi Din [679]

The average time that it takes for the car to travel the first 0.25m is 2.23 s

The average time that it takes for the car to travel the first 0.25 m is given by:

$t=\frac{2.24 s+2.21 s+2.23 s}{3}=2.227 s \sim 2.23 s$

The average time to travel just between 0.25 m and 0.50 m is 0.90 s

First of all, we need to calculate the time the car takes in each trial to travel between 0.25 m and 0.50 m:

$t_1 = 3.16 s - 2.24 s=0.92 s\\t_2 = 3.08 s- 2.21 s=0.87 s\\t_3 =3.15 s- 2.23 s=0.92 s$

Then, the average time can be calculated as

$t=\frac{0.92 s+0.87 s+0.92 s}{3}=0.90 s$

Given the time taken to travel the second 0.25 m section, the velocity would be 0.28 m/s

The velocity of the car while travelling the second 0.25 m section is equal to the distance covered (0.25 m) divided by the average time (0.90 s):

$v=\frac{d}{t}=\frac{0.25 m}{0.90 s}=0.28 m/s$

8 0

3 years ago

Read 2 more answers

What is the definition of dynamics in physics

grin007 [14]

A part of mechanics dealing with the movement of bodies under the activity of power.

5 0

3 years ago