You may have to______

A tire-pressure monitoring system warns you with a dashboard alert when one of your car tires is significantly under-inflated.

chubhunter [2.5K]

Answer:

The answers to the question are

(a) T = 2·π·r/v

(b) 3.3 % change in period of the under-inflated tire compared to the properly inflated tire

(c) Therefore the distance this car would have to travel in order for under-inflated tire to make one more complete rotation than the inflated tire is 57.685 meters.

Explanation:

(a) The period T = 2π/ω

The velocity v = ωr or ω = v/r

Therefore T = 2π/(v/r) = 2πr/v

T = 2·π·r/v

(b) Period of properly inflated tire with radius = 303 mm is 2π303/v

Period of under-inflated tire with radius = 293 mm is 2π293/v

Therefore we have percentage change in period of of the under-inflated tire compared to the properly inflated tire is given by

(2π303/v -2π293/v)/(2π303/v) = 2π10/v/(2π303/v) = 10/303 × 100 = 3.3 %

(c) The period of the under-inflated tire is 10/303 less than that of the inflated tire. Therefore for the under-inflated tire to make one complete turn more than the inflated tire, we have 1/(10/303) = 303/10 or 30.3 revolutions of either tire which is 30.3×2×π×303 = 57685.296 mm = 57.685 meters

Therefore the distance this car would have to travel in order for under-inflated tire to make one more complete rotation than the inflated tire is 57.685 meters

At 30.3 revolutions the distance covered by the under-inflated

= 55781.49 mm

Subtracting the two distances gives

1903.805 mm

The circumference of the inflated tire = 2×π×303 = 1903.805 mm

3 0

3 years ago

What’s the difference between engineering stress and strain and true stress and strain

Nana76 [90]

True strain and engineering strain? True stress is defined as the load divided by the cross-sectional area of the specimen at that instant and is a true indication of the internal pressures. ... Engineering stress is defined as the load divided by the initial cross-sectional area of the specimenAnswer:

Explanation:

4 0

3 years ago

Read 2 more answers

Calculate the electroosmotic velocity of an aqueous solution through a glass capillary 5 cm long with a 0.5 mm internal diameter

natita [175]

Answer:

Electroosmotic velocity will be equal to $1.6\times 10^{-4}m/sec$

Explanation:

We have given applied voltage v = 100 volt

Length of capillary L = 5 mm = 0.005 m

Zeta potential of the capillary surface $\xi =80mV=0.08volt$

Dielectric constant of glass is between 5 to 10 here we are taking dielectric constant as $\epsilon =10$

Viscosity of glass is $\eta =10^8$

Electroosmotic velocity is given as $v_{eo}=\frac{\epsilon \xi }{\eta }\times \frac{v}{L}$

$v_{eo}=\frac{10\times 0.08 }{10^8 }\times \frac{100}{0.005}=1.6\times 10^{-4}m/sec$

So Electroosmotic velocity will be equal to $1.6\times 10^{-4}m/sec$

8 0

3 years ago

Consider two different versions of algorithm for finding gcd of two numbers (as given below), Estimate how many times faster it

juin [17]

Answer:

Explanation:

Step 1:

a) The formula for compute greatest advisor is

gcd(m,n) = gcd (n,m mod n)

the gcd(31415,14142) by applying Euclid's algorithm is

gcd(31,415,14,142) =gcd(14,142,3,131)

=gcd=(3,131, 1,618)

=gcd(1,618, 1,513)

=gcd(1,513, 105)

=gcd(105, 43)

=gcd(43, 19)

=gcd(19, 5)

=gcd(5, 4)

=gcd(4, 1)

=gcd(1, 0)

=1

STEP 2

b) The number of comparison of given input with the algorithm based on checking consecutive integers and Euclid's algorithm is

The number of division using Euclid's algorithm =10 from part (a)

The consecutive integer checking algorithm:

The number of iterations =14,142 and 1 or 2 division of iteration.

14,142 ∠= number of division∠ = 2*14,142

Euclid's algorithm is faster by at least 14,142/10 =1400 times

At most 2*14,142/10 =2800 times.

5 0

3 years ago

Match the following items with their correct description.

Lera25 [3.4K]

Answer:

A. Manufacturers rating capacity ↔ 3. Must be marked on all jacks; must not be exceeded

B. Block Used to lift and hold heavy loads, allow them for travel ↔ 1. Place the jack head against this

C. Level surface ↔ 4. Place this under the base of the jack when it's necessary to provide a firm foundation

D. Jack ↔ 2. Used to lift and hold heavy loads, allow them for travel

Explanation:

The manufacturers rating for a jack is labelled on all jacks and should be referenced to compare with the load to be lifted so as to ensure a safe and successful lifting.

In order to lift a load, such as a car, it is required to place the jack on a level surface to provide balance during the lifting task

The head of the jack is placed against the block for lifting heavy objects for proper performance

8 0

3 years ago

You may have to_______