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Nadusha1986 [10]
3 years ago
13

A stunt driver rounds a banked, circular curve. The driver rounds the curve at a high, constant speed, such that the car is just

on the verge of skidding to the outside of the curve. A front view of a car driving on a banked curve. The cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car. Which forces are directly responsible for producing the car’s centripetal acceleration? Coriolis force centripetal force frictional force normal force gravitational force
Physics
1 answer:
bagirrra123 [75]3 years ago
5 0

Answer: C

Frictional force

Explanation:

The description of the question above is an example of a circular motion.

For a car travelling in a curved path, the frictional force between the tyres and the road surface will provide the centripetal force.

Since the road is banked, and the cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car, for cornering the banked road, the car will not rely only on the frictional force.

Therefore, the correct answer is option C - the frictional force.

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Is mixture an example of water
Evgesh-ka [11]
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1. explain why mid-sized cars and SUVs have different rollover fatality rates.
iris [78.8K]

Answer:

1.  They generally have poorer fuel efficiency and require more resources to manufacture than smaller vehicles, thus contributing more to climate change and environmental degradation. Their higher center of gravity increases their risk of rollovers.

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8 0
3 years ago
Acceleration toward the center of a curved path is called
Serggg [28]

Answer:

Centripetal acceleration.

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The force which causes this acceleration is centripetal force which is also directed towards the center of the circle and pulls the body towards its center.

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7 0
3 years ago
A 200.0 g block rests on a frictionless, horizontal surface. It is pressed against a horizontal spring with spring constant 4500
kenny6666 [7]

Answer:

 6 m/s

Explanation:

Given that :

mass of the block   m =  200.0 g  = 200 × 10⁻³ kg

the horizontal spring constant   k  =  4500.0 N/m

position of the block (distance x) = 4.00 cm  = 0.04 m

To determine the speed the block will be traveling when it leaves the spring; we applying the  work done on the spring as it is stretched (or compressed) with the kinetic energy.

i.e \frac{1}{2} kx^2  = \frac{1}{2} mv^2

kx^2 = mv^2

4500* 0.04^2 = 200*10^{-3} *v^2

7.2 =200*10^{-3}*v^{2}

v^{2}   =\frac{7.2}{200*10^{-3}}

v   =\sqrt{\frac{7.2}{200*10^{-3}}}

v = 6 m/s

Hence,the speed the block will be traveling when it leaves the spring is  6 m/s

5 0
3 years ago
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