Light and Reflection
Diagram Skills
E
STI
500
Mirrot
Flat Mirrors
1. The point of a 20.0 cm
D
pencil is placed 25.0 cm
from a flat mirror. Its
eraser is 15.0 cm from
the mirror. Three of the
light rays from the
pencil's point hit the
mirror with incident
angles of 0°, 20°, and
50° at points A, B, and C as shown.
a. Use a protractor to draw the reflected rays from points A, B, and C.
b. Where do reflected rays or their extensions intersect?
Mirror
B
c. What is the distance between the pencil's head and its image?
d. Would a person's eye located at point D perceive one of the reflected rays
drew? Will the person be able to see the image? Explain.
e. What if the eye is located at point E?
f. Draw incident rays from the eraser of the pencil to point A and to poin
Answers
The law of reflection allows to find the results for the questions about ray reflection in a plane mirror are:
a) Attachment we see a diagram of the incident and reflected rays, incident and reflected angles are equal.
b) The extension of the reflected rays is what forms the image.
c) The image's distance is 20 cm behind the flat mirror.
d) The point D (normal for an angle of 50º) cannot perceive the rays coming from point A, B, C
e) the Rays at points A, B, C cannot perceive in the point E.
f) attachment we see the rays that come out from the pencil eraser.
g) The image is behind the mirror at 15 cm.
The geometric interaction describes the interaction of light rays with surfaces, looking for where the rays are directed, it is described by two phenomenological laws:
Refraction. Establishes a relationship between incident rays and those transmitted by material means. Reflection. It establishes that the angle of incidence and reflection of the rays is the same.[tex]\theta_i = \theta_{r}[/tex]
From these two general laws, geometric optics establishes a relationship for the formation of the image, called the constructor's equation.
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
Where f is the focal length, p and q are the distance to the object and the image, respectively.
In this exercise, the medium is a mirror, which is why it must comply with the law of reflection.
a) In the attachment we see a diagram of the incident and reflected rays for the three points.
According to the law of reflection, the incident and reflected angles are equal.
b) From the diagram we can see that the extension of the reflected rays is what forms the image, which is called virtual and is located behind the mirror.
c) In the diagram we see two rays to form the image, we see that the distance to the object is equal to the distance to the image.
From the constructor's equation a plane mirror has an infinite radius.
p = -q
Therefore the image's distance is 20 cm behind the flat mirror. Therefore the distance to the object and the image are the same, the negative sign indicates that the image is behind the mirror.
d) A person located at point D (normal for an angle of 50º) cannot perceive the rays coming from point A, B, C since their angle of reflection is not equal to the incident angle.
To perceive a ray it must have an angle of incidence of 25º.
e) Point E is located very far from the pencil, so the incident angle increases as does the reflected angle.
the Rays at points A, B, C cannot perceive.
f) In the attachment we see the rays that come out from the pencil eraser, they indicate that the distance to the plane mirror is 15.0 cm,
g) The image is behind the mirror at 15 cm.
In conclusion using the law of reflection we can find the results for the questions are:
a) Attachment we see a diagram of the incident and reflected rays, incident and reflected angles are equal.
b) The extension of the reflected rays is what forms the image.
c) The image's distance is 20 cm behind the flat mirror.
d) The point D (normal for an angle of 50º) cannot perceive the rays coming from point A, B, C
e) the Rays at points A, B, C cannot perceive in the point E.
f) attachment we see the rays that come out from the pencil eraser.
g) The image is behind the mirror at 15 cm.
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Related Questions
Please help me.............................
Answers
Answer:
[tex]a[/tex]
Explanation:
is a corect anser
48.36
g.
MgSO4 to motes
Answers
Answer:120.3676
Explanation: using the molecular calculator and molar mass of MgSO4. hope this helps!
Dagmar says that diffusion happens really quickly. Is he right or wrong? Explain.
Answers
Answer:
Diffusion in gases is quick because the particles in a gas move quickly. It happens even faster in hot gases because the particles of gas move faster.
A pendulum with a length of 2 m has a period of 2.8 s. What is the period of a pendulum with a length of 8 m
Answers
Answer:
P = 2 pi (L / g)^1/2
P2 / P1 = (8 / 2)^1/2 = 2
The period would be twice as long or 5.6 sec.
One potato plant makes 7.5
potatoes. There are 10.000
people on a ship and 3
potatoes per person. How
many potato plants are
needed per person.
Answers
If all conditions are ideal, you may harvest about five to 10 potatoes per plant for your gardening efforts. Yields are based on both the care your give your plants during the growing season and the variety of potatoes you choose to grow.
answer
7.5×10.000=75÷3=25
An object is projected with speed of 4ms at an angle of 60° to horizontal. Calculate the time of flight of the object. (g=10ms2)
Answers
0.8 seconds
Explanation:
time of flight = 2u/g
u=4m/s
g=10
= 8/10
= 0.8 sec
just a trial...not sure!!!
Given :
∅ = 60⁰
u = 4 m/s
g = 10m/s²
to find :
T = ?
Solution :
as per formula,
[tex]t = \frac{2u \: sin \theta}{g} [/tex]
now put the value : [tex]t \: = \frac{2 \times 4 \times sin \: 60}{10} [/tex]
as we know [tex] sin60 \: = \frac{ \sqrt{3} }{2} [/tex]
therefore,
[tex]t \: = \frac{8 \times \frac{ \sqrt{3} }{2} }{10} [/tex]
as we solve this we get,
[tex]t \: = \frac{ 2\sqrt{3} }{5} [/tex]
that's t = 0.69 sec
[tex]\sf\fbox\red{\:I \:hope \:it's \:helpful \:to \:you}[/tex]
Understanding what motivates anyone is not easy because each individual has different
Answers
An irregularly shaped object weighs 11.20 N in air. When immersed in water, the object has an apparent weight of 3.83 N. Find its density.
Answers
Answer:
Weight of object = 11.2 N
Apparent weight = 3.83 N when immersed
Weight of water displaced = 11.2 - 3.83 = 7.37 N
d (density) W / V weight / volume the weight density
Wo = Vo do weight of object
Ww = Vo dw where Ww is weight of equivalent volume of water = 7.37
Wo / Ww = do / dw dividing previous equations
do = 11.2 / 7.37 dw = 1.52 dw
The density of the object is 1.52 that of water
The density of water is 1000 kg / m^3 * 9.8 m/s^2 = 9800 N/m^3
So the weight density is 14900 N/m^3
An irregularly shaped object weighs 11.20 N in air. When immersed in water, the object has an apparent weight of 3.83 N. It's density can be calculated as 1523 kg/m³.
To find the density, the given values are,
Weight in air = 11.20 N
Weight in water = 3.83 N
density of water = 1000 kg/m³
What is meant by Density?According to the Archimedes principle, when a body is immersed in a liquid partly or wholly, it experiences an upward force which is called buoyant force. The buoyant force is equal to the loss in weight of the body.
Loss in weight of the object = Weight of object in air - weight of object in water
Loss in weight = 11.20 - 3.83 = 7.37 N
Volume of body x density of water x g = 7.37
Let V be the volume of body
V x 1000 x 9.8 =7.37
V = 7.5× 10⁻⁴ m³
Weight in air = Volume of body x density of body x g
11.20 = 7.5× 10⁻⁴ x d x 9.8
d = 1523 kg/m³.
Thus, the density of body is 1523 kg/m³.
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A body is thrown up into the air takes a time of 4s to reach the height. What is the velocity with which the body was thrown up.(g=10ms2)
Answers
Answer:
40m/s
Explanation:
V= u + at
v= 0
a= -10
t= 4
0= u -40
u= 40m/s
just trial!!!!!!!
What is an ellipse?
a plane that slices between orbits
an oval-shaped orbit
a circular orbit
the center of gravity between orbiting objects
Answers
Answer:
i think it's C thx correct me if wrong
A circular disk of radius 0.200 m rotates at a constant angular speed of 2.50 rev/s. What is the centripetal acceleration (in m/s2) of a point on the edge of the disk?
Answers
[tex]a_c = 3.14\:\text{m/s}^2[/tex]
Explanation:
First, we need to convert the given angular speed [tex]\omega[/tex] from rev/s to rad/s:
[tex]2.50\:\dfrac{\text{rev}}{\text{s}}×\dfrac{2\pi\:\text{rad}}{1\:\text{rev}} = 15.7\:\text{rad/s}[/tex]
The centripetal acceleration [tex]a_c[/tex] is defined as
[tex]a_c = \dfrac{v^2}{r}[/tex]
Recall that [tex]v = r\omega[/tex] so we can write [tex]a_c[/tex] as
[tex]a_c = \dfrac{(r\omega)^2}{r} = \omega^2r[/tex]
[tex]\;\;\;\;\;=(15.7\:\text{rad/s})^2(0.200\:\text{m}) = 3.14\:\text{m/s}^2[/tex]
if the momentum of a 1,400 kg car is the same as the truck in question 17, what is the velocity of the car?
Answers
Answer:
Explanation:
momentum is mass times velocity
p = mv
so take the momentum of the truck in question 17 and divide by the mass of this car
v = p/m = p / 1400
in a compoumd are atoms physically or chemically combined
Answers
Answer:
They are...if I'm correct Chemically combined, sorry if I'm wrong.
If the velocity and frequency of a wave are both doubled, how does the wavelength change?
Answers
The wavelength will remain unchanged.
Explanation:
The velocity [tex]v[/tex] of a wave in terms of its wavelength [tex]\lambda[/tex] and frequency [tex]\nu[/tex] is
[tex]v = \lambda\nu[/tex] (1)
so if we double both the velocity and the frequency, the equation above becomes
[tex]2v = \lambda(2\nu)[/tex] (2)
Solving for the wavelength from Eqn(2), we get
[tex]\lambda = \dfrac{2v}{2\nu} = \dfrac{v}{\nu}[/tex]
We would have gotten the same result had we used Eqn(1) instead.
Answer:
the wavelength increases
Explanation:
is it true that playing badmenton help you to become a better person?
Answers
Answer:
There is no scientific evidence that playing specifically l badminton makes you a better person, but sport and exercise in general release hormones which can make you feel more happy therefore making you nicer to the people around you and 'a better person'.
Answer:
It's true because playing any sport makes a person happy. So a happy person is a better person.
Please Mark as brainliest.
I need help. please look at the image below and let me know I need this by 7:20 am pst.
Answers
Answer:
3(1.5) = 4.5 V
Explanation:
A 300 cm rope under a tension of 120 N is set into oscillation. The mass density of the rope is 120 g/cm. What is the frequency of the first harmonic mode (m
Answers
Answer:
Explanation:
f = [tex]\sqrt{T/(m/L)} / 2L[/tex]
T = 120 N
L = 3.00 m
(m/L) = 120 g/cm(100 cm/m / 1000 g/kg) = 12 kg/m
(wow that's massive for a "rope")
f = [tex]\sqrt{120/12} /(2(3))[/tex])
f = [tex]\sqrt{10\\}[/tex]/6 = 0.527 Hz
This is a completely silly exercise unless this "rope" is in space somewhere as the weight of the rope (353 N on earth) far exceeds the tension applied.
A much more reasonable linear density would be 120 g/m resulting in a frequency of √1000/6 = 5.27 Hz on a rope that weighs only 3.5 N
The mass of a car is 625kg. Calculate the weight of the gravitational field strength is 10 N/kg.
Answers
Ans: 62.5
Explanation: [tex]F{net}[/tex] = m x a
1N = 1kg x 1m/ [tex]s{2}[/tex]
A punter wants to kick a football so that the football has a total flight time of 4.70s and lands 56.0m away (measured along the ground). Neglect drag and the initial height of the football.
How long does the football need to rise?
What height will the football reach?
With what speed does the punter need to kick the football?
At what angle (θ), with the horizontal, does the punter need to kick the football?
Answers
Answer:
Explanation:
How long does the football need to rise?
4.70/3 = 2.35 s
What height will the football reach?
h = ½(9.81)2.35² = 27.1 m
With what speed does the punter need to kick the football?
vy = g•t = 9.81(2.35) = 23.1 m/s
vx = d/t = 56.0/4.70 = 11.9 m/s
v = √(vx²+vy²) = 26.0 m/s
At what angle (θ), with the horizontal, does the punter need to kick the football?
θ = arctan(vy/vx) = 62.7°
You are angry at Dr. Anderson for this exam, so you throw a 0.30-kg stone at his car with a speed of 44 m/s. How much kinetic energy does the stone have
Answers
Answer:
Explanation:
KE = ½mv²
KE = ½(0.30)44²
KE = 290 J rounded to 2 s.d.
What is the approximate value of k when 30 = e^5k?
Answers
Answer:
Explanation:
30 = e^5k
ln30 = lne^5k
ln30 = 5k
k = ln30/5
k = 0.68023947...
round to your heart's content.
A 40-kg worker climbs a ladder upwards for 15m. What work was done during their climb upwards?
Answers
Answer:
Explanation:
The work increased the potential energy
W = PE = mgh = 40(9.8)(15) = 5880 J(oules)
A 30-cm-tall, 4.0-cm-diameter plastic tube has a sealed bottom. 250 g of lead pellets are poured into the bottom of the tube, whose mass is 30 g, then the tube is lowered into a liquid. The tube floats with 5.0 cm extending above the surface. What is the density of the liquid
Answers
The density of the liquid will be equal to [tex]\rho=0.892 \ \dfrac{g}{cm^3}[/tex]
What is density?The density of an object is defined as the ratio of the mass of an object to the volume of the object.
Volume of tube = 2^2 * pi * 30 = 377 cm^3
Volume of tube submerged = 25* 377 / 30 = 314 cm^3
Buoyancy = weight of liquid displaced
Volume of liquid displaced = 314 cm^3
Mass of tube and lead = 250 + 30 = 280 g
Now from the mass density by definition
[tex]\rho = \dfrac{m}{v}[/tex]
[tex]m=\rho \times v[/tex]
Mass of liquid displaced = Mass being supported
[tex]314 \times \rho = 280 g[/tex]
[tex]\rho= \dfrac{280}{ 314 } = .892 \frac{g}{cm^3}[/tex]
Thus the density of the liquid will be equal to [tex]\rho=0.892 \ \dfrac{g}{cm^3}[/tex]
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The density of the liquid is 1.67 g/[tex]cm^3[/tex].
The volume of the tube is
30 * 4 * 3.14 * 0.25 = 94.2 [tex]cm^3[/tex].
The mass of the lead pellets and the plastic tube is
30 + 250 = 280 g.
The volume of the lead pellets is
250 / 11.34 = 22 [tex]cm^3[/tex].
The volume of the liquid that the tube displaces is
94.2 - 22 = 72 [tex]cm^3[/tex].
The density of the liquid is
280 / 72 = 1.67 g/ [tex]cm^3[/tex].
Therefore, the density of the liquid is 1.67 g/ [tex]cm^3[/tex].
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A car move at an initial velocity of 240m and reach at the final velocity of 540m in 8hours. calculate its acceleration.
Answers
Answer:
a = 0.01m/s²
Explanation:
V_f = V_0+a*t
V_f = Velocity final
V_0 = Velocity initial
a = acceleration
t = time
a = (V_f-V_0)/t
a = (540m/s-240m/s)/((8hr)*(60min/1hr)*(60s/1min))
a = 0.01m/s²
Help me outtttt jejjejejeje
Answers
Answer:
do it got a picture
Explanation:
a stone is thrown down off a bridge with a velocity of 22 m/s. what is its velocity after 1.5 seconds has passed?
Answers
Answer:
Velocity of the stone after 1.5 seconds has passed = 37 m/s
Explanation:
Initial velocity (u) = 22 m/s
Time (t) = 1.5 sec
Acceleration due to gravity (g) = 10 m/s²
By using kinematics equation:
v = u + gt
v = 22 + 10 × 1.5
v = 22 + 15
v = 37 m/s
Final velocity (v) = 37 m/s
what type of stretching is beneficial for sports performance and involves momentum
Answers
Answer:
Dynamic stretching
Explanation:
Dynamic stretching is a form of stretching beneficial in sports utilizing momentum from form.
Which of the following describes the motion of a block while it is in equilibrium? The block:
A. moves at a constant speed
B. slows down gradually to stop
C. speeds up for a bit, then moves at a constant speed
D. Accelerates constantly
Answers
The statement that describes the motion of a block while it is in equilibrium is: The block moves at a constant speed.
EQUILIBRIUM:A state of equilibrium in physics refers to a state of rest or the forces exerted on the object is in a balanced state.
In dynamic equilibrium, the acceleration of a body is zero. This means that the body is moving at a uniform speed.
Therefore, the statement that describes the motion of a block while it is in equilibrium is: The block moves at a constant speed.
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12) A horizontal force of 200 N is applied to move a 55-kg cart (initially at rest) across a 10 m level surface. What is the final speed of the cart? [hint: use work – energy principle] [3 marks]
Answers
Hi there!
We can use the following:
W = ΔKE = F · d
Find the work done on the cart:
W = 200 · 10 = 2000 J
Now, this is equal to the change in kinetic energy of the object. Its initial kinetic energy is 0 J since it starts from rest, so:
2000J = KEf - KEi
KE is given as:
[tex]KE = \frac{1}{2}mv^2[/tex]
2000J = 1/2(55)v²
4000 = 55v²
√(4000/55) = 8.53 m/s
3. a car takes off with initial velocity of 20m/s and move with uniform acceleration of 12m/s2 for 20s. It maintained a constant velocity for 30s and come to rest with uniform acceleration of 2m/s2. Calculate the total distance covered by the car.
Answers
The total distance traveled by the car at the given velocity and time is 900 m.
The given parameters:
initial velocity of the car, u = 20 m/sacceleration of the car, a = 12 m/s²time of motion of the car, t = 20 sfinal time = 30 sfinal acceleration = 2 m/s²The final time of motion of car before coming to rest is calculated as follows;
[tex]v_f = v_0 -at\\\\0 = 20 - 2t\\\\t = 10 \ s[/tex]
The graph of the car's motion is in the image uploaded.
The total distance traveled by the car is calculated as follows;
[tex]total \ distance = A \ + B \ + C\\\\total \ distance = (\frac{1 }{2} \times 20 \times 20) \ + (20 \times 30) \ + (\frac{1}{2}\times 10 \times 20)\\\\total \ distance = 900 \ m[/tex]
Thus, the total distance traveled by the car at the given velocity and time is 900 m.
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