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Nataly [62]
3 years ago
8

(20 points) Pressure is a measure of force per unit area or P=F/A. A student is checking the amount of pressure their car exerts

upon the road. The car weighs 2.25 tons ( US short ton ). The student finds the dimensions of each tire's footprint on the road is 12.5 cm X 16.5 cm. In pounds per square inch (psi), what is the total pressure the car exerts upon the road?
Physics
1 answer:
Gala2k [10]3 years ago
6 0

Answer:

35.2 ps

Explanation:

Hello!

A US short ton is equal to 2 000 pounds, therefore:

2.25 tons = 4500 pounds

Now lets convert the area of each tire footprint:

1 cm = 0.393701 in

12.5 cm = 4.92126 in

16.5 cm = 6.49607 in

Therefore, the area of the footprint is:   31.9688 in^2

Then, the pressure exerted by the car is given by the weigth devided by the total area, that is 4 times the area of the footprint :

4500 /(4 * 31.9688)  psi = 35.1906 psi

Rounding to the first tenth

35.2 psi

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Answer:

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Explanation:

To develop this problem it is necessary to apply the equations related to the Drag force and the Force of Gravity.

For the given point, that is, the moment at which the terminal velocity is reached, the two forces equalize, that is,

F_D =F_g

By definition we know that the Drag force is defined as

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Where,

C_d = Drag coefficient

\rho =Density

A =Cross-sectional Area

V = Velocity

In the other hand we have,

F_g = (m_1 +m_2) g

Where,

m_1 =Mass of sphere

m_2 =Mass of unknown object

Equating the two equations we have to

(m_1 +m_2) g=\frac{1}{2} C_d \rho A V^2

Re-arrange for m_2,

m_2 = \frac{1}{2g} C_d \rho A V^2 -m_1

Our values are given by,

C_d = 0.5

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V = 66.7m/s

m_1 = 3Kg

d= 32.7*10^{-2}m

r = 16.35*10^{-2}m

Replacing in the equation we have,

m_2 = \frac{1}{2(9.8)}(0.5) (1.22) (\pi*(16.35*10^{-2})^2)*66.7^2 -3

m_2 = 8.62Kg

<em>Therefore the mass of unknown object is 8.62Kg</em>

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3 years ago
A body undergoes SHM of amplitude 3cm and frequency 20Hz. What is the
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Answer:

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Explanation:

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Suppose 310. grams of ethanol (ethyl alcohol) is in an aluminum cup of 90.0 grams. Both of these are at 30.0C. A mass m of ice a
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Answer:

Explanation:

Given

mass of ethanol m_e=310\ gm

mass of aluminium cup m_{al}=90\ gm

both are at an initial temperature of T_i=30^{\circ}C

specific heat of ethanol c_e=2.46\ J/g-K

specific heat of aluminium c_{al}=0.9\ J/g-K

specific heat of ice c_i=2.108\ J/g-K

specific heat of water c_w=4.184\ J/g-K

Latent heat of fusion L=334\ J/gm

suppose m is the mass of ice added

Heat loss by Al cup and ethanol after 18^{\circ}C is reached

Q_1=(310\times 2.46+90\times 0.9)\cdot (30-18)

Heat gained by ice such that ice is melted and reached a temperature of 18^{\circ}C

Q_2=m\times 2.108\times (8.5)+m\times 334

Comparing 1 and 2 we get

m=23.65\ gm

Thus 23.65 gm of ice is added

                 

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