a sharp image is formed when light reflects from a

Answers

Answer 1

Answer:

Regular reflection

Explanation:

Regular reflection occurs when light reflects off a very smooth surface and forms a clear image.

i hope this helps a bit.

Answer 2

According to the context, a sharp image is formed when light reflects from a regular reflection.

What is regular reflection?

It is reflection without diffusion that obeys the laws of geometrical optics, as in mirrors.

This reflection of light happens when the angles that the two rays determine with the surface are equal.

Therefore, we can conclude that according to the context, a sharp image is formed when light reflects from a regular reflection.

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Related Questions

a cohesive force between the liquids molecules is responsible for the fluids is called​

Answers

Answer:

static force

Explanation:

mark me brainliest

In this experiment, you will use a track and a toy car to explore the concept of movement. You will measure the time it takes the car to travel certain distances, and then complete some calculations. In the space below, write a scientific question that you will answer by doing this experiment.​

Answers

Answer: if weight affects how fast they go?

Explanation:

Answer:

How can we change the speed of a toy car on a racetrack to describe the car’s motion?

Explanation:

thats the sample respond

A girl weighing 45kg is standing on the floor, exerting a downward force of 200N on the floor. The force exerted on her by the floor is ..............

Select one:

a.
No force exerted


b.
Less than 2000N


c.
Equal to 200 N


d.
Greater than 200 N​

Answers

Answer:

c.

Equal to 200 N..........

A cylindrical body has 6 m height and its radius is 2 metre calculate its volume. Ans :75.428m3​

Answers

Answer:

75.4

Explanation:

r= 2

h= 6

v= 22/7 *r*r*h

v= 75.42

Is there a way to see moon and the sun at once?

Answers

Yes there is a way to see the moon and sun at once when the sun is setting and the moon is rising

. Assume that the batter does hit the ball. If the bat's instantaneous angular velocity is 30 rad/s at the instant of contact, and the distance from the sweet spot on the bat to the axis of rotation is 1.25 m, what is the instantaneous linear velocity of the sweet spot at the instant of ball contact

Answers

Answer:

37.5 m/s

Explanation:

Using,

Formula

v = ωr....................... Equation 1

Where ω = instantaneous angular velocity, v = instantaneous linear velocity, r = radius or distance from the sweet spot of the bat to the axis of rotation.

From the question,

Given: ω = 30 rad/s, r = 1.25 m

Substitute these values into equation 1

v = 30(1.25)

v = 37.5 m/s.

Hence the instantaneous linear velocity of the sweet spot at the instant of ball contact is 37.5 m/s

Which of the following changes when an unbalanced force acts on an object?
A. mass
B. motion
C. inertia
D. weight

Answers

The answer is Motion

What is happening in the graph shown below?


A.
The object moves away from the origin at a speed of 3 m/s, stands still 6 m away from the origin for 3 seconds, then moves toward the origin at a speed of 2 m/s.
B.
The object moves toward the origin at a speed of 3 m/s, stands still 6 m away from the origin for 3 seconds, then moves away from the origin at a speed of 2 m/s.
C.
The object moves toward the origin at a speed of 6 m/s, stands still 6 m away from the origin for 3 seconds, then moves away from the origin at a speed of 8 m/s.
D.
The object moves away from the origin at a speed of 6 m/s, stands still 6 m away from the origin for 3 seconds, then moves toward the origin at a speed of 8 m/s.

Answers

Answer:

D. The object moves away from the origin at a speed of 3 m/s, stands still 6 m away from the origin for 3 seconds, then moves toward the origin at a speed of 2 m/s.

Explanation:

I just got it right lol

23
In order for a 12 Volt power source
to produce a current of 0.085 amps,
a resistance of...
[?] Ohms is needed.
Enter
Haven't learned this yet.

Answers

Answer:

141.18 ohms

Explanation:

From the question given above, the following data were obtained:

Voltage (V) = 12

Current (I) = 0.085 A

Resistance (R) =?

The resistance needed can be obtained as follow:

V = IR

12 = 0.085 × R

Divide both side by 0.085

R = 12 / 0.085

R = 141.18 ohms

Therefore, a resistor of resistance 141.18 ohms is needed.

A hair dryer draws a current of 12.8 A.

(a)How many minutes does it take for
6.8 x 10° C of charge to pass through the
hair dryer? The fundamental charge is
1.602 x 10-19 C.
Answer in units of min.

(b)How many electrons does this amount of
charge represent?
Answer in units of electrons.

Answers

Answer:

(a) 8.85×10⁻³ minutes

(b) 4.24×10¹⁹ electrons

Explanation:

(a) Using,

Q = it............................. Equation 1

Where Q = quantity of charge, i = current, t = time.

Make t the subject of the equation

t = Q/i............................. Equation 2

Given: Q = 6.8×10⁰ C, i = 12.8 A

Substitute these values into equation 2

t = 6.8×10⁰/12.8

t = 8.85×10⁻³ minutes

(b) n = Q/(1.602×10⁻¹⁹)................. Equation 3

Where n = number of electrons.

Given: Q = 6.8×10⁰ C

Substitute into equation 2

n = 6.8×10⁰/1.602×10⁻¹⁹

n = 4.24×10¹⁹ electrons

(a) The time taken by the charge to flow from the drier will be [tex]\d8.85[/tex]×[tex]\d10^{-3}[/tex]minutes

(b) Amount of the electrons in the charge will be  [tex]\d4.24[/tex]×[tex]\d10^{19}[/tex] electrons

What will be the time of the charge and number of the electrons in the charge ?

As we know Q = IT

Where Q = quantity of charge, i = current, T = time.

From the above equation

                    T= Q/I.

Given: Q = [tex]6.8\times\d10^{0}[/tex] C, i = 12.8 A

Substitute these values  

T=  [tex]6.8[/tex]×[tex]\d10^{0}[/tex] /12.8

T =  [tex]8.85[/tex]×[tex]\d10^{-3}[/tex] minutes

Now the number of the electrons present in the charge will be

n = Q/( [tex]1.602[/tex]×[tex]\d10^{-19}[/tex])

Where n = number of electrons.

Given: Q = [tex]6.8\times\d10^{0}[/tex] C

Substitute Value of Q  

n =  [tex]6.8\times\d10^{0}[/tex]/ [tex]1.602\times\d10^{-19}[/tex]

n = [tex]4.24\times\d10^{19}[/tex] electrons

Thus

(a)The time taken by the charge to flow from the drier will be [tex]\d8.85[/tex]×[tex]\d10^{-3}[/tex]minutes

(b)Amount of the electrons in the charge will be  [tex]\d4.24[/tex]×[tex]\d10^{19}[/tex] electrons

To know more about electric charge follow

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A light year is the amount of time it takes for light from the Sun to reach the Earth.
True
False

Answers

true! a light year is the amount of time it takes for light from the sun to reach Earth!

If you could help me please.
1) Does a 1 kg object weight 9.8 newtons on the moon? why?
2) How much does a 3-kg object weigh (on earth) in newtons?
3) How much does a 20-kg object weigh (on earth) in newton?
4) What must happen for the mass of an object to change?
5) What are 2 ways the weight of an object can change?

Answers

1) Does a 1 kg object weight 9.8 newtons on the moon? why?

No. 1kg of mass does not weigh 9.8N on the moon.

Weight = (mass) x (gravity).

Gravity is 9.8 m/s² on Earth, but gravity is only 1.62 m/s² on the moon.

2) How much does a 3-kg object weigh (on earth) in newtons?

Weight = (mass) x (gravity)

Gravity = 9.8 m/s² on Earth.

Weight = (3 kg) x (9.8 m/s² )

Weight = 29.4 N

3) How much does a 20-kg object weigh (on earth) in newton?

Weight = (mass) x (gravity)

Gravity = 9.8 m/s² on Earth.

Weight = (20 kg) x (9.8 m/s² )

Weight = 196 N

4) What must happen for the mass of an object to change?

When an object moves, its mass increases.  The faster it moves, the greater its mass gets.  But this is all part of Einstein's "Relativity".  The object has to move at a significant fraction of the speed of light before any change can be noticed or measured.  So as far as we are concerned, in everyday life, the mass of an object doesn't change, no matter where it is, or what you do to it.

5) What are 2 ways the weight of an object can change?

First, remember that the mass of an object doesn't change, no matter where it is, what you do to it, or what else is around it.

But its weight can change, because its weight depends on the strength of gravity in the place where the object is, and that gravity is the result of what else is around it in the neighborhood.  So the weight can change even though the mass doesn't.

The weight of an object changes if you take it to a place where gravity is stronger or weaker.

Let's say we have an object whose mass is 90.72 kilograms.  Like me !    

As long as I stay on earth, where gravity is 9.8 m/s² , I weigh 889 Newtons  (200 pounds).

. . . Fly me to the moon. Gravity = 1.62 m/s²  Weight = 147 Newtons (33 lbs)

. . . Drag me to Jupiter.  Gravity = 24.8 m/s²  Weight = 2,249 N (506 pounds)

My mass never changed, but my weight sure did.

Sam moves an 800 N wheelbarrow 5 meters in 15 seconds. How much work did he do?

Answers

Answer:

work done= force × displacement

=800×5

=4000J

Explanation:

The amount of work done is the result of the magnitude of force applied and the displacement of the body due to the force applied. Therefore, work done is defined as the product of the applied force and the displacement of the body.

a pendulum clock having Copper keeps time at 20 degree Celsius it gains 15 second per day if cooled to 0°C celsius calculate the coefficient of linear expansion of copper.​

Answers

?.............................

The amount of light that enters the pupil is controlled by the:
retina.
lens.
inis.

Answers

Answer: The amount of light that enters the pupil is controlled by the Iris

Explanation:

Define Mechanical advantage
fe effort of 2125N is used to lift a Lead of 500N
through a Verticle high of 2.N using a buly System
if the distance Moved by the effort is 45m
Calculate 1. Work done on the load
2. work done by the effort
3. Efficiency of the System​

Answers

Answer:

1) 1000Nm

2)  95,625Nm

3) 1.05%

Explanation:

Mechanical Advantage is the ratio of the load  to the effort applied to an object.

MA = Load/Effort

1) Workdone on the load = Force(Load) * distance covered by the load

Workdone on the load = 500N * 2m

Workdone on the load = 1000Nm

2)  work done by the effort = Effort * distance moves d by effort

work done by the effort = 2125 * 45

work done by the effort = 95,625Nm

3) Efficiency = Workdone on the load/ work done by the effort * 100

Efficiency = 1000/95625 * 100

Efficiency = 1.05%

Hence the efficiency of the system is 1.05%

At the base of a hill, a 90 kg cart drives at 13 m/s toward it then lifts off the accelerator pedal). If the cart just barely makes it to the top of this hill and stops, how high must the hill be?

Answers

Answer:

8.45 m

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 90 Kg

Initial velocity (u) = 13 m/s

Final velocity (v) = 0 m/s

Height (h) =?

NOTE: Acceleration due to gravity (g) = 10 m/s²

The height of the hill can be obtained as follow:

v² = u² – 2gh (since the cart is going against gravity)

0² = 13² – (2 × 10 × h)

0 = 169 – 20h

Rearrange

20h = 169

Divide both side by 20

h = 169/20

h = 8.45 m

Therefore, the height of the hill is 8.45 m

The gravitational force between two objects with masses 1kg and 28kg separated by a distance 7m is ____________10-11 N.


a.
3.81

b.
26.68

c.
9151.24

d.
1307.32
Hhhhhellllppp fastt​

Answers

Answer:

a. 3.81

Explanation:

F = GMm/r^2

F = (6.67 x 10^-11 x 28 x 1) / 7^2

F = 3.81 x 10^-11 N

1. A perspex box has a 10 cm square base and contains water to a height of 10 cm. A piece of rock of mass 600g is lowered into the water and the level rises to 12 cm.
(a) What is the volume of water displaced by the rock?
(b) What is the volume of the rock?
(c) Calculate the density of the rock

Answers

Answer:

(a) The volume of water is 100 cm³

(b) The volume of the rock is 20 cm³

(c) The density of the rock is 30 g/cm³

Explanation:

The given parameters of the perspex box are;

The area of the base of the box, A = 10 cm²

The initial level of water in the box, h₁ = 10 cm

The mass of the rock placed in the box, m = 600 g

The final level of water in the box, h₂ = 12 cm

(a) The volume of water in the box, 'V', is given as follows;

V = A × h₁

∴ The volume of water in the box, V = 10 cm² × 10 cm = 100 cm³

The volume of water in the box, V = 100 cm³

(b) When the rock is placed in the box the total volume, [tex]V_T[/tex], is given by the sum of the rock, [tex]V_r[/tex], and the  water, V, is given as follows;

[tex]V_T[/tex] = [tex]V_r[/tex] + V

[tex]V_T[/tex] = A × h₂

∴ [tex]V_T[/tex] = 10 cm² × 12 cm = 120 cm³

The total volume, [tex]V_T[/tex] = 120 cm³

The volume of the rock, [tex]V_r[/tex] = [tex]V_T[/tex] - V

∴ [tex]V_r[/tex] = 120 cm³ - 100 cm³ = 20 cm³

The volume of the rock, [tex]V_r[/tex] = 20 cm³

(c) The density of the rock, ρ = (Mass of the rock, m)/(The volume of the rock)

∴ The density of the rock, ρ = 600 g/(20 cm³) = 30 g/cm³

why food cook faster with salt water than cook with pure water​

Answers

Answer:

oil heats faster

Explanation:

When salt is added, it makes it harder for the water molecules to escape from the pot and enter the gas phase, which happens when water boils

If you blow across the open end of a soda bottle and produce a tone of 250 Hz, what will be the frequency of the next harmonic heard if you blow much harder?

___Hz

Answers

Answer:

Generally, the lowest overtone for a pipe open at one end and closed would be at  y / 4  where y represents lambda, the wavelength.

Since F (frequency) = c / y       Speed/wavelength

F2 / F1 = y1 / y2      because c is the same in both cases

F2 = y1/y2 * F1

F2 = 3 F1 = 750 /sec

Note that L = y1 / 4 = 3 y2 / 4 for these wavelengths to fit in the pipe

and y1 = 3 y2

The second harmonic will be three times the first harmonic. The answer is 750 Hz

VIBRATION OF WAVES IN PIPES

Closed pipes have odd multiples of frequencies or harmonics. That is,

If  [tex]F_{0}[/tex] = fundamental frequency = first harmonic

[tex]F_{1}[/tex] = 3[tex]F_{0}[/tex] = second harmonic

[tex]F_{2}[/tex] = 5[tex]F_{0}[/tex] = third harmonic

[tex]F_{3}[/tex] = 7[tex]F_{0}[/tex] = fourth harmonic

Let assume that the first harmonic is 250 Hz, If you blow it much harder, second, third or fourth harmonic can be produced.

By using the formula above,

second harmonic will be 3 x 250 = 750Hz

Therefore, the frequency of the next harmonic heard if you blow much harder will be 750 Hz

Learn more about Sound waves here: https://brainly.com/question/1199084

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