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

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

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.

Answer:

i think it's C thx correct me if wrong

Answer:

Explanation:

The work increased the potential energy

W = PE = mgh = 40(9.8)(15) = 5880 J(oules)

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

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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Answer:

[tex]a[/tex]

Explanation:

is a corect anser

Ans: 62.5

Explanation: [tex]F{net}[/tex] = m x a

                    1N = 1kg x 1m/ [tex]s{2}[/tex]

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:

Answer:

They are...if I'm correct Chemically combined, sorry if I'm wrong.

Answer:

Dynamic stretching

Explanation:

Dynamic stretching is a form of stretching beneficial in sports utilizing momentum from form.

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

Answer:

Explanation:

KE = ½mv²

KE = ½(0.30)44²

KE = 290 J                 rounded to 2 s.d.

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.

Answer:

do it got a picture

Explanation:

Answer:

3(1.5) = 4.5 V

Explanation:

Answer:

40m/s

Explanation:

V= u + at

v= 0

a= -10

t= 4

0= u -40

u= 40m/s

just trial!!!!!!!

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

The density of the liquid will be equal to   [tex]\rho=0.892 \ \dfrac{g}{cm^3}[/tex]

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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Answer:

Explanation:

30 = e^5k

ln30 = lne^5k

ln30 = 5k

k = ln30/5

k = 0.68023947...

round to your heart's content.

[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]

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.

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

Answer:120.3676

Explanation: using the molecular calculator and molar mass of MgSO4. hope this helps!

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

The statement that describes the motion of a block while it is in equilibrium is: The block moves at a constant speed.

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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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]

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°

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

The total distance traveled by the car at the given velocity and time is 900 m.

The given parameters:

The 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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