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3 years ago

PLEASE HELP ME WHATS THE ASNWRR

Physics

2 answers:

stealth61 [152]3 years ago

7 0

Answer:

think its b i dunno

Explanation:

AysviL [449]3 years ago

6 0

The answer is b that is the only option that makes sense

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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 is Nuclear energy

Yuliya22 [10]

Answer:

Nuclear energy is that released by dividing the nuclei of heavy atoms (also called under the name Nuclear Fission)

Explanation:

In this process, a large amount of heat is generated that can be used to obtain mechanical energy, which is used to generate electrical energy.

Nuclear energy is a sustainable source of energy having a low impact on the environment.

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3 years ago

How does the mass-energy equation e = mc2 relate to fission

Strike441 [17]

It relates because<span> the </span>energy<span> of an object to its </span>mass and since fission requires energy to do so, e=mc^2 relates to it.

3 0

4 years ago

Read 2 more answers

An infinite plane of charge has surface charge density 7.2 Î¼C/m^2. How far apart are the equipotential surfaces whose potential

Rina8888 [55]

Answer:

so the distance between two points are

$d = 0.246 \times 10^{-3} m$

Explanation:

Surface charge density of the charged plane is given as

$\sigma = 7.2 \mu C/m^2$

now we have electric field due to charged planed is given as

$E = \frac{\sigma}{2\epsilon_0}$

now we have

$E = \frac{7.2 \times 10^{-6}}{2(8.85 \times 10^{-12})}$

$E = 4.07 \times 10^5 N/C$

now for the potential difference of 100 Volts we can have the relation as

$E.d = \Delta V$

$4.07 \times 10^5 (d) = 100$

$d = \frac{100}{4.07 \times 10^5}$

$d = 0.246 \times 10^{-3} m$

3 0

3 years ago

What is the maximum force that could be applied to anterior cruciate ligament (ACL) if it has a diameter of 4.8 mm and a tensile

vovangra [49]

Answer:

Maximum force, F = 1809.55 N

Explanation:

Given that,

Diameter of the anterior cruciate ligament, d = 4.8 mm

Radius, r = 2.4 mm

The tensile strength of the anterior cruciate ligament, $P=100\times 10^6\ N/m^2=10^8\ Pa$

We need to find the maximum force that could be applied to anterior cruciate ligament. We know that the unit of tensile strength is Pa. It must be a type of pressure. So,

$F=P\times A\\\\F=10^8\times \pi (2.4\times 10^{-3})^2\\\\F=1809.55\ N$

So, the maximum force that could be applied to anterior cruciate ligament is 1809.55 N

4 0

4 years ago