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3 years ago

What remains the same in a compound regardless of the quantity?

1 answer:

8 0

The ratio btw number of atoms.
example: water H2O is the compound. No matter how much water you take (a liter, 10 cubic meters... whole swimming pool), the proportion btw hydrogen atoms and oxygen would be always 2 to 1.

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Which of the following is an example of uniform circular motion?

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Answer:

C. A rocket traveling to the Moon

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3 years ago

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Fred points a toy laser gun at a wall. Considering that the frequency of the gun's light is 4.91 × 1014 hertz and that Planck’s

Stella [2.4K]

The answer is 3.24 x 10^-21

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3 years ago

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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.

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3 years ago

What type of energy is released when a firecracker explodes

Ludmilka [50]

-- Sound energy. That's how you hear it explode.

-- Light energy. That's how you see it in the sky.

-- Heat energy. That's why you get burned if sparks fall on you.

-- Potential energy, as all the pieces fall to the ground.

These were all stored in it as chemical energy before it exploded.

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4 years ago

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Please help on this one!!

inysia [295]

The energy transformations that occur as you coast down long hill on a bicycle, including the brakes to make the bike stop at the bottom, is that at the top of the hill you have high GPE AND LOW KE, on your way down you have HIGH KE AND LOW GPE, and at the bottom you have thermal energy due to the stop of the brakes.

the law of conversation of energy and describe the energy transformations that occur as you coast down a long hill on a bicycle and then apply the brakes to make the bike stop at the bottom.

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3 years ago