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3 years ago

"where are the majority of transform faults located?"

1 answer:

5 0

<span>The majority of the transform faults in the ocean are located on the ocean floor. They offset oceanic ridges. However, not all transform faults are on the oceanic crust- some are found on continental margins. Transform faults are boundaries between plates of the Earth's crust.</span>

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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

2 years ago

Kristen is running a half marathon. By her tenth minute of running, her breathing rate has increased from her resting rate.

pashok25 [27]

They need more oxygen to function, so gas exchange needs to increase

7 0

2 years ago

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The gravitational force between two objects is represented by the variable F. If the distance is decreased by 5 what is the new

Anvisha [2.4K]

The universal law of gravitation states that:

Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of distance between them.

It means that if the gravitational force is F, then if the distance is decreased by 5 times, then the new gravitation force is:

F/5² = F/25

6 0

2 years ago

Any help????????????

pashok25 [27]

The answer should be D

3 0

2 years ago

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Why does the moon's gravity have a greater effect on the earth's ocean than the sun.

White raven [17]

Answer:

Tides on our planet are caused by the gravitational pull of the Moon and Sun. Earth's oceans "bulge out" because the Moon's gravity pulls a little harder on one side of our planet (the side closer to the Moon) than it does on the other. The Sun's gravity raises tides, too, but lunar tides are twice as big.

4 0

3 years ago