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Lemur [1.5K]
4 years ago
6

For a plane mirror,how is the object s related to the distance s?

Physics
1 answer:
dlinn [17]4 years ago
3 0
Light is a very complex phenomenon, but in many situations its behavior can be understood with a simple model based on rays and wave fronts. A ray is a thin beam of light that travels in a straight line. A wave front is the line (not necessarily straight) or surface connecting all the light that left a source at the same time. For a source like the Sun, rays radiate out in all directions; the wave fronts are spheres centered on the Sun. If the source is a long way away, the wave fronts can be treated as parallel lines.

Rays and wave fronts can generally be used to represent light when the light is interacting with objects that are much larger than the wavelength of light, which is about 500 nm. In particular, we'll use rays and wave fronts to analyze how light interacts with mirrors and lenses.

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My very eager mother just served us nine pizzas
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A vehicle travelling at an initial velocity of 20km/hr,accelerates at 4m/s².calculate its final velocity after 10 seconds.​
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acceleration = Velocity changes ÷ time of the velocity changes

4 m/s^2 =

4 × 10^(-3) × 3600 km / h =

4 × 3.6 =

14.4 km / h

Thus :

14.4 = V(2) - V(1) / t(2) - t(1)

14.4 = V(2) - 20 / 10

Multiply both sides by 10

10 × 14.4 = 10 × ( V(2) - 20 ) / 10

144 = V(2) - 20

Add both sides 20

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5 0
3 years ago
Read 2 more answers
When 597 J of heat are added to a gas, it expands. Its internal energy increases by 318 J. How much work does the gas do? (Unit=
damaskus [11]

The magnitude of work done by the gas is 279 J and the sign is negative so W = -279 J as work is done by the system.

<u>Explanation:</u>

According to first law of thermodynamics, the change in internal energy of the system is equal to the sum of the heat energy added or released from the system with the work done on or by the system. If the heat energy is added to the system to perform a certain work, then the heat energy is taken as positive, while it will be negative when the heat energy is released from the system.

Similarly, in this case, the heat energy of 597 J is added to the system. So the heat energy will be positive, while the gas expansion occurs means work is done by the system.

          ΔU = Q+W

Since ΔU is the change in internal energy which is given as 318 J and the heat energy added to the system is Q = 597 J.

Then the work done by the gas = ΔU - Q = 318 J - 597 J = - 279 J.

As the work is done by the system, so it will be denoted in negative sign and the magnitude of work done by the gas is 279 J.

7 0
3 years ago
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