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Volgvan
1 year ago
12

An engineer wants to design an oval racetrack such that 3.20 × 10 3 lb racecars can round the exactly 1000 ft radius turns at 10

2 mi/h without the aid of friction. She estimates that the cars will round the turns at a maximum of 175 mi/h. Find the banking angle θ necessary for the race cars to navigate the turns at 102 mi/h without the aid of friction.
Physics
1 answer:
Reptile [31]1 year ago
4 0

Answer:

The banking angle necessary for the race cars is 34.84°

Explanation:

For normal reaction the expression is:

\\Nsin\theta = \frac{mv^{2} }{R}  =Fc\\tan\theta =\frac{v^{2} }{Rg}  \\\theta =tan^{-1} (\frac{v^{2} }{Rg} )\\\theta =tan^{-1} (\frac{(102*0.447)^{2} }{1000*0.3048*9.8} )=34.84

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In 1993, Ileana Salvador of Italy walked 3.0 km in under 12.0 min. Suppose that during 115 m of her walk Salvador is observed to
dexar [7]

Answer:

t = 25 seconds

Explanation:

Given that,

Distance, d = 115 m

Initial speed, u = 4.2 m/s

Final speed, v = 5 m/s

We need to find the time taken in increasing the speed.

We know that,

Acceleration, a=\dfrac{v-u}{t} ....(1)

The third equation of kinematics is as follows :

v^2-u^2=2ad\\\\\text{Put the value of a in above equation}\\\\v^2-u^2=2\times \dfrac{v-u}{t}\times d\\\\\because (a^2-b^2)=(a-b)(a+b)\\\\(v-u)(v+u)=\dfrac{2\times (v-u)d}{t}\\\\t=\dfrac{2d}{v+u}\\\\\text{Putting all the values}\\\\t=\dfrac{2\times 115}{4.2+5}\\\\t=25\ s

Hence, it will take 25 seconds to increase the speed.

6 0
2 years ago
Process in which permanent deformation of metals occurs due to applied stress and results in breaking of bonds and then reformin
luda_lava [24]

Answer:

This process involves the motion of dislocations and is termed slip (or glide in some textbooks)

Explanation:

Plastic deformation of metals (and other crystalline materials) usually occurs by slip, which is the sliding of planes of atoms over one another by dislocation movements.

On a microscopic scale, stress causes planes of crystalline objects to leave their original position and slide over other planes into new positions, these microscopic movements manifest as a slip on a macroscopic scale. And the planes do not return back to their original position after the removal of the dislocation-causing stress.

6 0
2 years ago
How much force is required to drag a 90 lb. box up this "frictionless" inclined plane? 109 lb. 10 lb. 81 lb. 9 lb.
MrRissso [65]
I believe is 10 lb if not it's 9 lb.
5 0
2 years ago
Read 2 more answers
A chair of mass 30.0 kg is at rest on a horizontal floor. The floor is not frictionless. You push on the chair with a force of 8
miv72 [106K]
First make sure you draw a force diagram. You should have Fn going up, Fg going down, Ff going left and another Fn going diagonally down to the right. The angle of the diagonal Fn (we'll call it Fn2) is 35° and Fn2 itself is 80N. Fn2 can be divided into two forces: Fn2x which is horizontal, and Fn2y which is vertical. Right now we only care about Fn2y.

To solve for Fn2y we use what we're given and some trig. Drawing out the actual force of Fn2 along with Fn2x and Fn2y we can see it makes a right triangle, with 80 as the hypotenuse. We want to solve for Fn2y which is the opposite side, so Sin(35)=y/80. Fn2y= 80sin35 = 45.89N

Next we solve for Fg. To do this we use Fg= 9.8 * m. Mass = 30kg, so Fg = 9.8 * 30 = 294N.

Since the chair isn't moving up or down, we can set our equation equal to zero. The net force equation in the vertical direction will be Fn + Fn2y -Fg = 0. If we plug in what we know, we get Fn + 45.89 -294 = 0. Then solve this algebraically.

Fn +45.89 -294 = 0
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7 0
2 years ago
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A roundabout in a fairground requires an input power of 2.5 kW when operating at a constant angular velocity of 0.47 rad s–1 . (
natita [175]

Answer:

Explanation:

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= 5.32 kN . m

= 5 kN.m

b )

When power is switched off , it will decelerate because of frictional torque .

5 0
2 years ago
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