A car drives over the top of a hill that has a radius of 50 m .

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The phrase of centripetal acceleration is,

The primary principles needed to resolve this dilemma would be the Newton’s 2nd legislation and centripetal force.
Response just

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Step two of 2

The worth of normal force is zero, in the event that vehicle will travel from the ground. Therefore, don’t just take any non-zero value.

A car drives over the top of a hill that has a radius of 50 m. What maximum speed can the car have without flying off the road at the top of the hill? A car drives over the top of a hill that has a radius of 50 m. What maximum speed can the car have without flying off the road at the top of the hill? A car drives over the top of a hill that has a radius of 50 m. What maximum speed can the car have without flying off the road at the top of the hill? A car drives over the top of a hill that has a radius of 50 m. What maximum speed can the car have without flying off the road at the top of the hill? A car drives over the top of a hill that has a radius of 50 m. What maximum speed can the car have without flying off the road at the top of the hill? A car drives over the top of a hill that has a radius of 50 m. What maximum speed can the car have without flying off the road at the top of the hill? A car drives over the top of a hill that has a radius of 50 m. What maximum speed can the car have without flying off the road at the top of the hill? A car drives over the top of a hill that has a radius of 50 m. What maximum speed can the car have without flying off the road at the top of the hill? A car drives over the top of a hill that has a radius of 50 m. What maximum speed can the car have without flying off the road at the top of the hill? A car drives over the top of a hill that has a radius of 50 m. What maximum speed can the car have without flying off the road at the top of the hill? A car drives over the top of a hill that has a radius of 50 m. What maximum speed can the car have without flying off the road at the top of the hill?
x = gr
The phrase of maximum velocity is,

x = gr
The force is understood to be the item of mass of item and acceleration.
Right here, may be the centripetal acceleration, m may be the mass of item, v may be the velocity of item, and r may be the radius of circular course.

Vmax = M(9.8 m/s2)(50 m) = 22.1 m/s

The extra weight W of item is,

Right here, W may be the fat of item, m may be the mass of item, and g may be the acceleration as a result of gravity.
Replace 0 for N to get the maximum rate permitted.

Step by step

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9.8 m/s?
Rearrange the equation.

Right here, g may be the acceleration as a result of gravity and r may be the radius.
General guidance

At first, use the Newton’s 2nd legislation on system and equate the force as a result of gravity and centripetal force. Finally, rearrange the phrase the maximum velocity.

The utmost rate of vehicle is 22.1 m/s. The rate of vehicle needs to be under the utmost rate so the vehicle won’t travel from the road. In the event that rate surpasses this optimum value, then vehicle will travel from the road. The utmost rate of vehicle could be determined utilising the acceleration as a result of gravity and radius.

The utmost rate of vehicle is 22 m/s

The utmost rate of vehicle is 22 m/s

Response
Substitute 50 m for r and for g.

Basics

mg -N-my?
W = mg
The equation of movement is,

Ideas and explanation

The equation of movement the system of vehicle at the top of a mountain could be written utilising the Newton’s 2nd legislation. The phrase of maximum velocity of vehicle without traveling from the road towards the top of the mountain could be determined utilising the force as a result of gravity and centripetal force.

Right here, mg may be the fat of vehicle, N may be the normal force functioning on the vehicle through the ground, m may be the mass of vehicle, v may be the velocity of vehicle, and r may be the radius of circular course.

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