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Zolol [24]
3 years ago
13

Determine the number of revolutions through which a typical automobile tire turns in 1 yr. Suppose the automobile travels 13500

miles each year on tires with radius 0.220 m.
Physics
1 answer:
RSB [31]3 years ago
5 0

Answer:

Number of revolution made by tire is 1.57 x 10⁷

Explanation:

Radius of tire, r = 0.220 m

Circumference of tire, C = 2πr

Substitute the value of r in the above equation.

C = 2 x π x 0.220 m = 1.38 m

Total distance covered by tire in a year, D = 13500 miles

But 1 mile = 1609.34 m

So, D = 13500 x 1609.34 m

Number of revolutions take by tire, N = \frac{D}{C}

N=\frac{13500\times1609.34}{1.38}

N = 15743543

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3 years ago
An automobile steering wheel is shown. What is the ideal mechanical advantage? If the AMA is 8, what is the efficiency of the st
Mars2501 [29]

1. Ideal Mechanical Advantage (IMA): 9

Explanation:

For a wheel and axle system like the steering wheel, the IMA is given by:

IMA=\frac{r_w}{r_a}

where

r_w is the radius of the wheel

r_a is the radius of the axle

For the steering wheel of the problem, we see that r_w = 18 cm and r_a=2 cm, so the IMA is

IMA=\frac{18 cm}{2 cm}=9


2. Efficiency: 88.9%

Explanation:

The efficiency of a system is defined as the ratio between the AMA (actual mechanical advantage) and the IMA:

\eta=\frac{AMA}{IMA}\cdot 100

In this problem, AMA=8 and IMA=9, so the efficiency is

\eta=\frac{8}{9}\cdot 100=88.9\%


6 0
4 years ago
A block slides down a rough ramp with a 30-degree incline as shown.
In-s [12.5K]

Answer:

Image 4 ?

Explanation:

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4 years ago
What force is required to give an object with mass 2000 kg an acceleration of 3.5 m/s2
ch4aika [34]
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8 0
4 years ago
Read 2 more answers
What is the magnitude of a point charge in coulombs whose electric field 58 cm away has the magnitude 2.8 N/C?
Yuliya22 [10]

Answer:

q = 1.05 × 10^-10 C

Explanation:

E = 2.8 N/C

r = 58cm = 0.58m

k = 9×10^9

q = ?

E = f/q

E = kq/r²

q = E*r²/k

q = (2.8×0.58²)÷9×10^9

q = 1.05 × 10^-10 C

8 0
3 years ago
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