Question number 8 <br> Plz help

A 3,220 lb car enters an S-curve at A with a speed of 60 mi/hr with brakes applied to reduce the speed to 45 mi/hr at a uniform

Grace [21]

The magnitude of the total friction force exerted by the road on the tires at B will be 919.46 lb.

<h3>What is the friction force?</h3>

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. Its unit is Newton (N).

Mathematically, it is defined as the product of the coefficient of friction and normal reaction.

The given data in the problem is;

The weight is,W= 3,220 lb

The speed is,u= 60 mi/hr

The reducing speed is,v= 45 mi/hr

The distance traveled is,d= 300 ft

The radius of curvature of the path of the car at B is,R= 600 ft.

1 mile = 5280 ft

From the Newtons' equation of motion;

$\rm v^2 = u^2 +2ad \\\\ \rm (45 \times \frac{5280}{3600} )^2 = (60 \times \frac{5280}{3600} )^2 +2a\times 300 \\\\$

The tangential accelerations are;

$\rm a_t = \frac{66^2 -88^2}{600} \\\\ \rm a_t = -5.65 ft/sec^2 \\\\$

The force is found as;

$\rm \sum F = ma \\\\ \frac{3220}{32.2} \times -5.65 \\\\ F_T= 565 \ lb$

The normal force is;

$\rm F_n = \frac{3220}{32.2} \times \frac{66^2}{600} \\\\ F_N =726 \ lb$

The net or the total friction force exerted by the road on the tires at B. is found as;

$\rm F = \sqrt{(565)^2+(726)^2} \\\ F = 919.946 \ lb$

Hence, the magnitude of the total friction force exerted by the road on the tires at B will be 919.46 lb.

To learn more about the friction force, refer to the link;

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6 0

2 years ago

By what factor will the Electrostatic Force between two charged objects change when the amount of charge on both objects doubles

Mademuasel [1]

Answer:

F' = (4/9)F

Explanation:

The electrostatic force between two charged objects is given by Coulomb's Law:

F = kq₁q₂/r² -------------------- equation (1)

where,

F = Electrostatic Force

k = Coulomb's Constant

q₁ = magnitude of first charge

q₂ = magnitude of second charge

r = distance between charges

Now, when the charges and distance altered as follows:

q₁' = 2q₁

q₂' = 2q₂

r' = 3r

Then,

F' = kq₁'q₂'/r'²

F' = k(2q₁)(2q₂)/(3r)²

F' = (4/9)kq₁q₂/r²

using equation (1):

7 0

3 years ago

PLEASE NEED THIS QUICKLY

saw5 [17]

The potential energy of the block is A) 490 J

Explanation:

The potential energy of an object is the energy possessed by the object due to its position in the gravitational field.

It is calculated as follows:

$PE=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height of the object above the ground

For the block in this problem, we have:

m = 10 kg

$g=9.8 m/s^2$

h = 5 m

Therefore, its potential energy is:

$PE=(10)(9.8)(5)=490 J$

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

3 0

4 years ago

You are on a ParKour course. First you climb a angled wall up 9.5 meters. They you shimmy along the edge of a 3.5 meter long wal

Svet_ta [14]

The average speed would be 0.65 m/s, therefore the correct option is (A).

The average speed is calculated by the formula

Average speed= (total distance/ total time)

The total distance of the trip=9.5+3.5+15=28 m

The total time of trip=43 sec

Therefore the average speed=28/43=0.65 m/s.

8 0

4 years ago

Wha is amplitde in sound

NemiM [27]

Answer:

The number of molecules displaced in a vibration makes the amplitude of a sound.

7 0

3 years ago

Question number 8 Plz help