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Inga [223]
3 years ago
5

If a car takes a banked curve at less than a given speed, friction is needed to keep it from sliding toward the inside of the cu

rve (a real problem on icy mountain roads). (a) Calculate the minimum speed, in meters per second, required to take a 84 m radius curve banked at 16° so that you don't slide inwards, assuming there is no friction. (b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 16 km/h?

Physics
2 answers:
kvasek [131]3 years ago
7 0

Answer:

Please see attachment

Explanation:

Please see attachment

zubka84 [21]3 years ago
4 0

Answer:

minimum speed is 15.35 m/s

frictional coefficient  is 0.26

Explanation:

given data

radius = 84 m

angle = 16°

speed = 16 km/h = 4.43 m/s

to find out

minimum speed and   minimum coefficient

solution

we will apply here formula for velocity that is

velocity² = radius × g × tanθ

v² = 84 × 9.8 × tan16

v² =  236.04

v = 15.35 m/s

and

we find first friction force here

friction force 1 = m v² /r

friction force 1 = m (15.35)² / 84 = 2.80 m

and

friction force 2 = m v² /r

friction force 2 = m (4.43)² / 84 =  0.245 m

so total friction force = f1 - f2

total friction force = 2.80 - 0.245  = 2.55 m

so frictional coefficient = friction force /g

frictional coefficient = 2.55 / 9.8

so frictional coefficient  is 0.26

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Answer:

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We are given;

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Thus;

150 = ½×60(v_f² - 2²)

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Answer:

the third piece landed 2\,\sqrt{2}  meters in a direction 45 degrees from the observer's right towards the front

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Due to conservation of momentum, the momentum of the third piece must add to the other two momenta in vector from to render zero  (initial momentum of the firework on a plane parallel to ground. Therefore the momentum components of the third piece must have a component to the right equal in magnitude to the momentum of the piece that traveled to the left, and a momentum component pointing to the front of the observer of equal magnitude to that of the piece that traveled behind the observer. Then the direction of the third momentum must be at 45 degrees from the observer's right towards the observer's front.

Also, the magnitude of a momentum vector of such components, is given by the Pythagorean theorem. Recall also that the magnitudes of the momentum vectors of the first two pieces must be equal since they traveled equal distances.

|P_3|=\sqrt{P_1^2+P_2^2} \\|P_3|=\sqrt{P^2+P^2}\\|P_3|=\sqrt{2P^2}\\|P_3|=P\sqrt{2}

Then the distance traveled by the third piece must also keep this proportionality:

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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}}

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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

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x_{0} is initial position

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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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The answer would be:

A, a space station orbiting Earth.

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<h3><u>Answer;</u></h3>

Sedimentary rock

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<h3><u>Explanation</u>;</h3>
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