The reflection in a clear window of a store
is a(n)

Answers

Answer 1

The reflection in a clear window of a store is a(n) image.

Why are images seen as reflection?

Images are seen as reflections because of the behavior of light. When light strikes a smooth, reflective surface such as a mirror or still water, it bounces off the surface at the same angle at which it hit it. This process is called reflection. The reflected light rays then travel to our eyes, creating an image.

The angle of incidence (the angle at which the light strikes the surface) is equal to the angle of reflection (the angle at which the light bounces off the surface). This causes the reflected image to be a mirror image of the original object.

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Related Questions

Move numbers to the boxes to show the factor pairs of 14. Response area with 4 blank spaces Blank space 1 empty times Blank space 2 empty equals 14 Blank space 3 empty times Blank space 4 empty equals 14 Answer options with 14 options. Answer Options 1 2 3 4 5 6 7 8 9 10 11 12 13 1

Answers

Move numbers to the boxes to show the factor pairs of 14:

Blank space 1: 1

Blank space 2: 1

Blank space 3: 2

Blank space 4: 7

A factor pair of a number is a pair of whole numbers that can be multiplied together to give the original number. For the number 14, the factor pairs are (1,14) and (2,7). So, we can put 1 in the first blank, 14 in the second blank, 2 in the third blank, and 7 in the fourth blank to show the factor pairs of 14.

To show the factor pairs of 14 in the given response area with 4 blank spaces, we need to find the two numbers that can be multiplied together to give 14. These two numbers are called factor pairs of 14.

To begin, we can start listing the factors of 14. The factors of 14 are 1, 2, 7, and 14. We can then use these factors to form factor pairs by multiplying them together. The factor pairs of 14 are (1, 14) and (2, 7).

To show these factor pairs in the given response area, we can put the first factor of each pair in the first and third blank spaces, and the second factor of each pair in the second and fourth blank spaces. Therefore, we can put 1 in the first blank, 14 in the second blank, 2 in the third blank, and 7 in the fourth blank to show the factor pairs of 14.

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The side of a cube of metal is measured to be (1.00±0.06) cm and its mass is measured to be (41.0±0.4) g. Determine the uncertainty in the density of the solid in kilograms per cubic meter.

Answers

The density of the solid is (4.10 ± 0.78) × 10^3 kg/m^3.

To calculate the density of the cube below formula can be used:

ρ = m/V

where ρ is density, m is mass, and V is volume. For a cube, the volume is given by:

V = (side)^3

Therefore, the uncertainty in density can be calculated using the formula:

δρ/ρ = sqrt[(δm/m)^2 + 3(δs/s)^2]

where δρ is the uncertainty in density, δm is the uncertainty in mass, δs is the uncertainty in side, and s is the value of the side.

Now, putting in the given values:

s = (1.00 ± 0.06) cm = 0.01 ± 0.0006 m

m = (41.0 ± 0.4) g = 0.0410 ± 0.0004 kg

Volume, V = (0.01 m)^3

                  = 1.0 × 10^-6 m^3

Therefore, the density is:

ρ = m/V

  = 0.0410 kg/1.0 × 10^-6 m^3

  = 4.10 × 10^4 kg/m^3

Now substituting the values and calculating the uncertainty in density:

δρ/ρ = sqrt[(δm/m)^2 + 3(δs/s)^2]

δρ/ρ = sqrt[(0.0004/0.0410)^2 + 3(0.0006/0.01)^2]

δρ/ρ = 0.019

Therefore, the uncertainty in density is:

δρ = (0.019)(4.10 × 10^4 kg/m^3)

     = 779 kg/m^3

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Work is done on a wooden crate by pushing it across a floor. The work done is equal to the force applied parallel to the floor multiplied by the

A. force of friction
B. distance moved
C. mass of the crate
D. crate's direction

Answers

The work done is equal to the force applied parallel to the floor multiplied by the distance moves. Hence, option B is correct.

When a wooden crate is pushed across a floor, the work done on it is given by the product of the force applied parallel to the floor and the distance moved by the crate.

This is because work is defined as the product of force and displacement in the direction of the force, which in this case is the distance moved by the crate. The force of friction and the mass of the crate are not relevant to the calculation of work done in this scenario

It is important to understand the concept of work and the factors that influence it, as it is a fundamental concept in physics and is used in many real-world applications.

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3.
A steel container (the coefficient of linear expansion = 10-5 °C-1) with a volume
of 6 liters filled with acetone (the coefficient of volume expansion = 1.5 x 10-
3 °C-1). If the container and acetone are heated from 0 °C to 40 °C, what is the
volume of spilled acetone?
(6)

Answers

The amount of acetone that was spilt is around 0.36 liters, or 360 milliliters.

How to determine volume?

To solve this problem, use the formula for volumetric thermal expansion:

ΔV = V₀βΔT

Where:

ΔV = change in volume

V₀ = initial volume

β = coefficient of volumetric expansion

ΔT = change in temperature

Also use the formula for linear thermal expansion to find the change in length of the container:

ΔL = L₀αΔT

Where:

ΔL = change in length

L₀ = initial length

α = coefficient of linear expansion

ΔT = change in temperature

Initial length can be calculated as follows:

V = L³ ⇒ L = ∛V = ∛6 L ≈ 1.82 meters

Now calculate the change in length of the container and the change in volume of acetone:

ΔL = L₀αΔT = (1.82 m)(10⁻⁵ °C⁻¹)(40 °C) ≈ 0.00073 meters

ΔV = V₀βΔT = (6 liters)(1.5 x 10⁻³ °C⁻¹)(40 °C) ≈ 0.36 liters

Since the acetone spills out of the container, its final volume is equal to the initial volume minus the change in volume:

Vf = Vi - ΔV = 6 L - 0.36 L = 5.64 L

Therefore, the volume of spilled acetone is approximately 0.36 liters or 360 milliliters.

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