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2 years ago

7

A friend asks you how much pressure is in your car tires. You know that the tire manufacturer recommends 30 psi, but it's been a

while since you've checked. You can't find a tire gauge in the car, but you do find the owner's manual and a ruler. Fortunately, you've just finished taking physics, so you tell your friend, "I don't know, but I can figure it out." From the owner's manual you find that the car's mass is 1000 kg . It seems reasonable to assume that each tire supports one-fourth of the weight. With the ruler you find that the tires are 15 cm wide and the flattened segment of the tire in contact with the road is 14 cm long. What answer will you give your friend?

1 answer:

sergiy2304 [10]2 years ago

7 0

you have 28 psi in the tire

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A block of gelatin is 120mm by 120mm by 40mm whrn unstressed. A force of 49N is applied tangentially to the upper surface causin

Inessa05 [86]

Answer:

The shearing stress is 10208.3333 Pa

The shearing strain is 0.25

The shear modulus is 40833.3332 Pa

Explanation:

Given:

Block of gelatin of 120 mm x 120 mm by 40 mm

F = force = 49 N

Displacement = 10 mm

Questions: Find the shear modulus, Sm = ?, shearing stress, Ss = ?, shearing strain, SS = ?

The shearing stress is defined as the force applied to the block over the projected area, first, it is necessary to calculate the area:

A = 40*120 = 4800 mm² = 0.0048 m²

The shearing stress:

$Ss=\frac{F}{A} =\frac{49}{0.0048} =10208.3333Pa$

The shearing strain is defined as the tangent of the displacement that the block over its length:

$SS=tan\theta =\frac{Displacement}{L} =\frac{10}{40} =0.25$

Finally, the shear modulus is the division of the shearing stress over the shearing strain:

$Sm=\frac{10208.3333}{0.25} =40833.3332Pa$

6 0

2 years ago

A magnetic field of 37.2 t has been achieved at the mit francis bitter national magnetic laboratory. Find the current needed to

jeka57 [31]

Answer:

Here is the complete question:

https://www.chegg.com/homework-help/questions-and-answers/magnetic-field-372-t-achieved-mit-francis-bitter-national-magnetic-laboratory-find-current-q900632

a) Current for long straight wire $=3.7\ MA$

b) Current at the center of the circular coil $=2.48\times 10^{5}\ A$

c) Current near the center of a solenoid $236.8\ A$

Explanation:

⇒ Magnetic Field due to long straight wire is given by (B),where $B=\frac{\mu\times I}{2\pi(r) },so\ I=\frac{B\ 2\pi(r)}{\mu}$

$\mu=4\pi \times 10^{-7}\ \frac{henry}{m}$

Plugging the values,

Conversion $1\times 10^6 A = 1\ MA$ ,and $2cm=\frac{2}{100}=0.02\ m$

$I=\frac{37.2\times \ 2\pi(0.02)}{4\ \pi \times (10^{-7})}=3.7\ MA$

⇒Magnetic Field at the center due to circular coil (at center) is given by, $B=\frac{\mu\times I (N)}{2(a)}$

So $I= \frac{2B(a)}{\mu\ N} = \frac{2\times 37.2\times 0.42}{4\pi\times 10^{-7}\times 100}=2.48\time 10{^5}\ A$

⇒Magnetic field due to the long solenoid, $B=\mu\ nI=\mu (\frac{N}{l})I$

Then $I=\frac{B}{\mu(\frac{N}{L})} \approx 236.8\A$

So the value of current are $3.7 MA$ , $2.48\times 10^{5} A$ and $236.8\ A$ respectively.

8 0

2 years ago

A 3250 N car is pushed a distance of 35 m the power was 11375 J, how long did it take?

irakobra [83]

Answer:

10.8s

Explanation:

Given parameters:

Force on the car = 3250N

Distance = 35m

Power = 11375W

Unknown:

Time taken = ?

Solution:

To solve this problem;

Power is the rate at which work is done

Power = $\frac{work done }{time}$

Work done = force x distance = 3250 x 35 = 123200J

Now;

11375 = $\frac{123200}{t}$

11375t = 123200

t = 10.8s

5 0

2 years ago

Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

3 0

2 years ago

1. A mass is on a level plane, it has a weight of 20N. What is the coefficient of kinetic friction if an applied force

Arturiano [62]

Answer:

0.4

Explanation:

F-Fr=ma where F is applied force, Fr is friction, m is mass and a is acceleration.

Since the mass is moving with a constant velocity, there's no acceleration hence

$F=Fr=\mu N$ where N is the weight of object and \mu is coefficient of kinetic friction.

$F=\mu N and making [tex]\mu$ the subject

$\mu=\frac {F}{N}$

Substituting F for 8 N and N for 20 N

$\mu=\frac {8}{20}=0.4$

Therefore, coefficient of kinetic friction is 0.4

4 0

3 years ago