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4 years ago

A 175-kg motorcycle goes around an unbanked turn of radius 75.4 m at a steady 112 km/h. Determine its acceleration?

Physics

1 answer:

Nuetrik [128]4 years ago

7 0

Answer: $12.83 m/s^{2}$

Explanation:

In this case we are talking about the centripetal acceleration $a_{c}$ , which can be calculated by:

$a_{c}=\frac{V^{2}}{r}$

Where:

$V=112 \frac{km}{h} \frac{1000 m}{1 km} \frac{1h}{3600 s}=31.11 m/s$ is the speed of the motorcycle

$r=75.4 m$ is the radius of the path the motorcycle is going around

Solving:

$a_{c}=\frac{(31.11 m.s)^{2}}{75.4 m}$

Finally:

$a_{c}=12.83 m/s^{2}$

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3 years ago

Ford used to make a 7.0 liter engine. Calculate how many many cubic inches would be the same size engine.

Anestetic [448]

Answer : The volume of engine in cubic inches is, $427.166\text{ inch}^3$

Explanation :

As we are given that the volume of engine is, 7.0 liters. Now we have to calculate the volume of engine in cubic inches.

Conversion used :

$1\text{ liter}=1000cm^3$

So, $7\text{ liter}=7000cm^3$

$1\text{ inch}=2.54cm$

or,

$1cm=\frac{1}{2.54}\text{ inch}$

$1cm^3=(\frac{1}{2.54}\text{ inch})^3$

As,

$1cm^3=(\frac{1}{2.54}\text{ inch})^3$

So,

$7000cm^3=\frac{7000cm^3}{1cm^3}\times (\frac{1}{2.54}\text{ inch})^3=427.166\text{ inch}^3$

Thu, the volume of engine in cubic inches is, $427.166\text{ inch}^3$

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3 years ago

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

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Whitepunk [10]

B; to produce a hypothesis.

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3 years ago

Read 2 more answers

a body is projected at an angle of 30 degrees to the horizontal with a velocity of 15m/s, calculate the time it takes to reach t

andreev551 [17]

Explanation:

The vertical component the velocity of the projectile is 15 m/s x sin 30 = 7.5 m/s.

The body is accelarating downwards at 10 m/s^2.

This means that every second its upward velocity reduces by 10 m/s.

So if the body is travelling upwards at 7.5 m/s then how long does it take for the velocity to become 0?

(7.5 m/s) / (10 m/s^2) = 0.75 s

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3 years ago