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11Alexandr11 [23.1K]
2 years ago
9

Experiment to show that light travels in a straight line​

Physics
1 answer:
SCORPION-xisa [38]2 years ago
6 0
The answer here
Two experiments are used to demonstrate how light travels in straight lines. In the first example, the presenter arranges three pieces of card, with holes in, in an uneven line. The light stops and cannot travel through all three cards. When she arranges the holes in a straight line, the light can travel through.
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Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc
4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance D_{A} beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed v_{A}. Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed v_{B}, which is greater than v_{A}.

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: t=\frac{D_{A}}{v_{B}-v_{A}}

              Part B: x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

x=x_{0}+vt

where

x_{0} is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is D_{A}.

The equation will be:

x_{A}=D_{A}+v_{A}t

Car B started at the starting line. So, its equation is

x_{B}=v_{B}t

Part A: When they meet, both car are at "the same position":

D_{A}+v_{A}t=v_{B}t

v_{B}t-v_{A}t=D_{A}

t(v_{B}-v_{A})=D_{A}

t=\frac{D_{A}}{v_{B}-v_{A}}

Car B meet with Car A after t=\frac{D_{A}}{v_{B}-v_{A}} units of time.

Part B: With the meeting time, we can determine the position they will be:

x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )

x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Since Car B started at the starting line, the distance Car B will be when it passes Car A is x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}} units of distance.

5 0
3 years ago
BRAINLIEST AND MAY DOUBLE POINTS!!!
raketka [301]

Given that:

 Energy of bulb  (Work ) = 30 J,

 Time (t) = 3 sec

The power consumption =  ?

 We know that, Power can be defined as rate of doing work

              Power (P) = Work(Energy supplied) ÷ time

                               = 30 ÷ 3

                              = 10 Watts

<em> The power consumption is 10 W.</em>


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Coronary artery disease is a disease of the heart where the arteries and blood vessels become clogged with fat deposits called _
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Plaque is the reasoning for this question ;)
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Suppose the acceleration of the particle moving gin a circle of radius "r" with the uniform speed "v" is proportional to some po
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So i say the power or the radius cortex in my equinox when i look at this V is the determination of V
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What is equal to the kinetic energy of a car with a mass of 0.5t (tonne) if it travels evenly at 70 km/h?
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Formula for kinetic energy is 1/2mv^2 so that answer should most probably be 94521.6J
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