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When changing a tire sized 195/75 R15 to a 5 percent lower profile, the correct tire size would be _____

valentinak56 [21]

Answer and explanation:

When you are changing a car tire, the most important thing is to keep the total diameter as equal as possible.

The total car tire diameter can be calculated as:

$D_{tot orig}=\frac{195 \cdot 75}{2540 \cdot 2}+15''=26.5''$

The profile of this tire is 75 (the higher/taller relation), therefore a 5 percent lower profile would be:

pr=0.95·75=71.25

The problem is that the profiles are normalized and the nearest profile available is 70.

If we take a theorical tire with a profile of 71.25:

$D_{tot orig}=D_{new}\\\frac{195 \cdot 75}{2540 \cdot 2}+15''=\frac{X \cdot 71.25}{2540 \cdot 2}+15''\\X=205.26$

The theorical tire size should be 205/71 R15.

If we look for a real tire size, we should look for a tire with a diameter nearest to 26.5'' and a profile of 70.

The best option for real tire size is: Tire 225/70 R14 (wheel diameter of 26.4'') or 205/70 R15 (wheel diameter of 26.3'').

3 0

4 years ago

. What is the velocity of a free-<br> falling object after 5 seconds?<br> (Use 10 m/s2 for gravity.)

Viefleur [7K]

Answer:

vf = 50 m/s

Explanation:

The equation for this kinematic problem is:

vf = vi + at

We are given:

a = 10m/s^2

vi = 0m/s

t = 5 sec

vf = ?

Solve for final velocity:

vf = 0 + 10(5)

vf = 50 m/s

8 0

3 years ago

Fine grains of beach sand are assumed to be spheres of radius 64.8 µm. These grains are made of silicon dioxide which has a dens

Nutka1998 [239]

Answer:

1) Mass of the grain is $2.9632\times 10^{-9} kg$ .

2) 0.08901 kg of sand would have surface area equal the surface area of the cube.

Explanation:

1) Radius of the grain,r = 64.8 µm = $6.48\times 10^{-5}m$

Volume of the sphere = $\frac{4}{3}\pi r^3$

Volume of the grain of a sand:

$V=\frac{4}{3}\times \pi r^3=\frac{4}{3}\times 3.14\times (6.48\times 10^{-5} m)^3$

$V=1.1397\times 10^{-12} m^3$

Density of a grain of sand = $d=2600 kg/m^3$

Mass of a grain of a sand = M

$d=2600 kg/m^3=\frac{M}{1.1397\times 10^{-12} m^3}$

$M=2.9632\times 10^{-9} kg$

Mass of the grain is $2.9632\times 10^{-9} kg$ .

2) Surface are of sphere: $4\pi r^2$

Surface area of a grain:

$A=4\times 3.14\times (6.48\times 10^{-5}m)^2$

$A=5.2766\times 10^{-8} m^2$

Length of the cube = a = 0.514 m

Total surface area of cube ,A'= $6a^2$

$A'=6\times (0.514 m)^2=1.5851 m^2$

let the number grains with area equal to total surface area of cube be x.

$A'=A\times x$

$x=\frac{1.5851 m^2}{5.2766\times 10^{-8} m^2}=3.003\times 10^7$

Volume of x number of grains :V'

$V'=V\times x$

$V= 1.1397\times 10^{-12} m^3\times 3.003\times 10^7$

$V'=3.4236\times 10^{-5} m^3$

Mass of $3.4236\times 10^{-5} m^3$ of sandL:

= $3.4236\times 10^{-5} m^3\times 2600 kg/m^3=0.08901 kg$

0.08901 kg of sand would have surface area equal the surface area of the cube.

8 0

3 years ago

There are 8.8 liters of gas in a piston at a pressure of 1.75 atmospheres. The temperature remains constant, and the gas is comp

saveliy_v [14]

The new pressure P2 is 2.48 atmosphere.

<u>Explanation:</u>

Here, the one of the product of pressure and volume is equal to the products of pressure and volume of other.

By using Boyles's law,

pressure is inversely proportional to volume,

P1 V1 = P2 V2

where P1, V1 represents the first pressure and volume,

P2, V2 represents the second pressure and volume

P2 = (P1 V1) / V2

= (1.75 $\times$ 8.8) / 6.2

P2 = 2.48 atmosphere.

5 0

3 years ago

Lee knew that certain events occur to create new material in the process of sea-floor spreading. Rearrange the steps into the co

Marina86 [1]

Answer:

Seafloor spreading is a geologic process in which tectonic plates—large slabs of Earth's lithosphere—split apart from each other. ... As tectonic plates slowly move away from each other, heat from the mantle's convection currents makes the crust more plastic and less dense.

Explanation:

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8 0

3 years ago