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mote1985 [20]
3 years ago
6

A level curve on a country road has a radius of 150 m. What is the maximum speed at which this curve can be safely negotiated on

a rainy day when the coefficient of friction between the tires on a car and the road is 0.40?
Physics
1 answer:
SSSSS [86.1K]3 years ago
8 0

The maximum speed of the car is 24.3 m/s

Explanation:

For a car moving along an unbanked turn, the frictional force provides the centripetal force required to keep the car in circular motion. Therefore, we can write:

\mu mg = m\frac{v^2}{r}

where the term on the left is the frictional force while the term on the right is the centripetal force, and where

\mu=0.40 is the coefficient of friction between the tires and the road

m is the mass of the car

g=9.8 m/s^2 is the acceleration of gravity

v is the speed of the car

r = 150 m is the radius of the curve

Solving for v, we find the (maximum) speed at which the car can move along the turn:

v=\sqrt{\mu gr}=\sqrt{(0.40)(9.8)(150)}=24.3 m/s

For speed larger than this value, the frictional force is no longer enough to keep the car along the turn.

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

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Uploaded imageTwo long, parallel wires each carry the same current I in the same direction (see figure below). The total magneti
LekaFEV [45]

Answer:

Zero

Explanation:

Two long parallel wires each carry the same current I in the same direction. The magnetic field in wire 1 is given by :

B_1=\dfrac{\mu_oI_1}{2\pi r}

Magnetic force acting in wire 2 due to 1 is given by :

F_{2}=I_2lB_1

F_2=\dfrac{\mu_oI_1I_2 l}{2\pi r}

Similarly, force acting in wire 1 is given by :

F_1=\dfrac{\mu_oI_1I_2 l}{2\pi r}According to third law of motion, the force acting in wire 1 will be in opposite direction to wire 2 as :

F_2=-F_1

So, the total magnetic field at the point P midway between the wires is in what direction will be zero as the the direction of forces are in opposite direction.  

7 0
3 years ago
Five race cars speed toward the finish line at the Jasper County Speedway. The table lists each car’s speed in meters/second. If
Anarel [89]
I think the answer would be Car C.
7 0
3 years ago
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If a vector that is 3cm long represents 30 km/h, what velocity does a 5 cm long vector which is drawn using the same scale repre
FrozenT [24]

Answer:

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8 0
3 years ago
Consider a situation of simple harmonic motion in which the distance between the endpoints is 2.39 m and exactly 8 cycles are co
aivan3 [116]

Answer:

1.195 m

2.8375 s

2.21433 rad/s

Explanation:

d = Distance = 2.39 m

N = Number of cycles = 8

t = Time to complete 8 cycles = 22.7 s

Radius would be equal to the distance divided by 2

r=\frac{d}{2}\\\Rightarrow r=\frac{2.39}{2}\\\Rightarrow r=1.195\ m

The radius is 1.195 m

Time period would be given by

T=\frac{t}{N}\\\Rightarrow T=\frac{22.7}{8}\\\Rightarrow T=2.8375\ s

Time period of the motion is 2.8375 s

Angular speed is given by

\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{2.8375}\\\Rightarrow \omega=2.21433\ rad/s

The angular speed of the motion is 2.21433 rad/s

4 0
4 years ago
- A fridge has a maximum static friction force
Dennis_Churaev [7]

Answer:

They will move the fridge if they all push in the same direction, but it will not move with constant velocity

Explanation:

The maximum static friction force is

F_f = -250 N (negative sign since its direction is opposite to the push applied by the people)

Sam can apply a force of 130 N, while Amir and Andre can apply a push of 65 N each, so the total force that they can apply, if they push in the same direction, will be:

F=130 + 65 +65=260 N

This force is larger than the frictional force, so the fridge will start moving.

However, the net force on the fridge will be:

\sum F = 260 N - 250 N = 10 N

And according to Newton's second law,

\sum F = ma

where m is the mass of the fridge and a its acceleration, since the net force is not zero, then the fridge will have a non-zero acceleration, so it will not move with constant  velocity.

3 0
4 years ago
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