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KIM [24]
3 years ago
6

Tires of a Bigfoot truck has a diameter of 2.2 m. If it rotates 60 revolutions find distance travel on the road.

Physics
1 answer:
Firlakuza [10]3 years ago
8 0

Answer:

s = 414.7 m\\

Explanation:

The relationship between the linear distance covered by an object and its angular displacement is given by the following formula:

s = rθ

where,

s = distance traveled on road = ?

r  radius of tires = diameter/2 = 2.2 m/2 = 1.1 m

θ = angular displacement = (60 rev)(2π rad/1 rev) = 377 rad

Therefore,

s = (1.1 m)(377 rad)\\s = 414.7 m

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About 3.88 meters per second

Explanation:

14km= 14,000m

14,000m per hour

so 14,000 ÷ 60= 233.33 meters per minute

233.33÷60=3.88 meters per second

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Wherever a river loses energy
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Derek runs 4 laps around the track. If each lap around the track is 0.25 miles long, and he starts and stops in the same locatio
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In a standard tensile test, a steel rod of 7 8-in. diameter is subjected to a tension force of 17 kips. Knowing that ν = 0.30 an
natali 33 [55]

Answer:

(a) Elongation of the rod==5.61×10⁻⁹m

(b) Change in diameter=1.640×10⁻⁸m

Explanation:

Given data

Diameter d=78 in=1.9812 m

Cross Area is:

A=(\pi /4)d^{2} \\A=(\pi /4)(1.9812m)^{2}\\A=3.08m^{2}

Applied Load P=17 KN=17×10³N

E=29 × 106 psi=1.99947961×10¹¹Pa

Stress and Strain in x direction

Stress in x direction

σ=P/A

=\frac{17*10^{3}N }{3.08m^{2} }\\ =5517.25Pa

σ=5517.25 Pa

Strain in x direction

ε=σ/E

=\frac{5517.25}{1.99947961*10^{11} } \\=2.76*10^{-8}

ε=2.76×10⁻⁸

Part (a)

Elongation of the rod=Lε

=(0.2032)(2.76×10⁻⁸)

Elongation of the rod==5.61×10⁻⁹m

Part(b) Change in diameter

Strain in y direction

ε₁= -vε

ε₁= -(0.30)(2.76×10⁻⁸)

ε₁=-8.28×10⁻⁹

Change in diameter=d×ε₁

Change in diameter=(1.9812m)×(-8.28×10⁻⁹)

Change in diameter=1.640×10⁻⁸m

4 0
3 years ago
2. How much do you think the force of friction must be? Why?
Tamiku [17]

Answer:

It must be high do to the gravity

Explanation:

4 0
2 years ago
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