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Brut [27]
3 years ago
12

Disturbed by speeding cars outside his workplace, Nobel laureate Arthur Holly Compton designed a speed bump (called the "Holly h

ump") and had it installed. Suppose a 1 800-kg car passes over a hump in a roadway that follows the arc of a circle of radius 18.8 m.
a. If the car travels at 33.0 km/h what force does the road exert on the car as the car passes the highest point of the hump?
b. What is the maximum speed the car can have without losing contact with the road as it passes this highest point?
Physics
1 answer:
Bezzdna [24]3 years ago
3 0
:<span>  </span><span>30.50 km/h = 30.50^3 m / 3600s = 8.47 m/s 

At the top of the circle the centripetal force (mv²/R) comes from the car's weight (mg) 

So, the net downward force from the car (Fn) = (weight - centripetal force) .. and by reaction this is the upward force provided by the road .. 

Fn = mg - mv²/R 
Fn = m(g - v²/R) .. .. 1800kg (9.80 - 8.47²/20.20) .. .. .. ►Fn = 11 247 N (upwards) 
(b) 
When the car's speed is such that all the weight is needed for the centripetal force .. then the net downward force (Fn), and the reaction from the road, becomes zero. 

ie .. mg = mv²/R .. .. v² = Rg .. .. 20.20m x 9.80 = 198.0(m/s)² 

►v = √198 = 14.0 m/s</span>
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Explanation:

It is given that,

The ramp is tilted upwards at 25 degrees and Paul is pulling a large crate up the ramp with a rope that angles 10° above the ramp.

Total angle with respect to ramp is 35 degrees.

If Paul pulls with a force of 550 N.

The horizontal component of the force is given by :

F_x=F\ cos\theta

F_x=550\ cos(35)

F_x=450.53\ N

The vertical component of the force is given by :

F_y=F\ sin\theta

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3 years ago
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Answer:

the object has least potential energy at mean position of the SHM

Explanation:

If a block is connected with a spring and there is no resistive force on the system

In this case the total energy of the system is always conserved and it will change from one form to another form

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3 years ago
What is the minimum force require to move a 5kg wooden crate on a wooden floor?
kolbaska11 [484]

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By Newton's second law, the horizontal and vertical net forces are

• net horizontal:

∑ <em>F</em> = <em>p</em> - <em>f</em> = 0

• net vertical:

∑ <em>F</em> = <em>n</em> - <em>w</em> = 0

where

<em>p</em> = magnitude of the <u>p</u>ushing force

<em>f</em> = mag. of <u>f</u>riction

<em>n</em> = mag. of the <u>n</u>ormal force

<em>w</em> = <u>w</u>eight of the crate

The second equation gives

<em>n</em> = <em>w</em> = (5 kg) (9.8 m/s²) = 49 N

Friction is proportional to the normal force by a factor of <em>µ</em>, so

<em>f</em> = <em>µ</em> (49 N) = 49<em>µ</em> N

To overcome static friction, the push has to exceed this in magnitude, so that

<em>p</em> > 49<em>µ</em> N

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Answer:

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