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choli [55]
3 years ago
10

A student is driving through a mountainous region where the road is at some times flat, at some times inclined upward, and at so

me time inclined downward. The student maintains a speed of 20 m/s on the roadway, but is required to make an emergency stop on the three sepearte occasions. On levels roadway, it takes 25 m to stop. On a downward-sloping roadway, it takes 40 m to stop. On an upward-sloping roadway, it takes 18 m to stop. Explain why the stopping distances are different. (Focus answer using work and energy, other concepts may be used as well but be sure work and energy are included.)
Physics
1 answer:
VLD [36.1K]3 years ago
7 0

Answer:

Explanation:

It is frictional force of the ground that helps in bringing the vehicle to stop . In the process of stopping , negative work is done on the car by friction force to overcome its kinetic energy .

At levelled road , for stoppage

Kinetic energy of vehicle = Work done by frictional force . = friction force x displacement .

At upward slopping road , gravitational force acting downward also helps the vehicle to stop do friction has to do less work .

At upward inclined  road , for stoppage

Kinetic energy of vehicle = Work done by frictional force + work done by gravitational force  = (friction force + gravitational force ) x displacement .

Hence displacement is less .

At downward slopping road ,  friction has to do more work because friction has to do work against gravitational force acting downwards wards and kinetic energy of the vehicle  also .

At downward inclined  road , for stoppage

Kinetic energy of vehicle + work done by gravitational force  = Work done by frictional force = friction force  x displacement .

Hence displacement is more .

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<u>Answer</u>

0.00346 hL

<u>Explanation</u>

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10,000 cL = 1 hL

∴ 34.6 cL = 34.6/10,000  hL

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Business are necessary in order to preform which of the following functions
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Hire, organize, and supply workers.

Explanation:

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Answer: the airy pattern can only arise from wave propagation

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2 years ago
If the coefficient of static friction is 0.357, and the same ladder makes a 58.0° angle with respect to the horizontal, how far
zavuch27 [327]

Answer: d= 0.57* l

Explanation:

We need to check that before ladder slips the length of ladder the painter can climb.

So we need to satisfy the equilibrium conditions.

So for ∑Fx=0, ∑Fy=0 and ∑M=0

We have,

At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal

At the top of ladder, N₂ acting horizontal

And Between somewhere we have the weight of painter acting downward equal to= mg

So, we have N₁=mg

and also mg*d*cosФ= N₂*l*sin∅

So,

d=\frac{N2}{mg}*l * tan∅

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So f₁= 0.357 * 69.1 * 9.8

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