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Amanda [17]
3 years ago
12

The pressure in car tires is often measured in pounds per square inch (lb/in.2lb/in.2), with the recommended pressure being in t

he range of 25 to 45 lb/in.2lb/in.2. Suppose a tire has a pressure of 25.5 lb/in.2lb/in.2 . Convert 25.5 lb/in.2lb/in.2 to its equivalent in atmospheres. Express the pressure numerically in atmospheres.
Physics
1 answer:
krok68 [10]3 years ago
8 0

Answer:

25.5 pounds per square inch are equivalent to 1.735 atmospheres.

Explanation:

An atmosphere equals 14.695 pounds per square inch. We find the equivalent of given pressure in atmospheres by means of simple rule of three:

x = 25.5\,\frac{lb}{in^{2}} \times \frac{1\,atm}{14.695\,\frac{lb}{in^{2}} }

x = 1.735\,atm

25.5 pounds per square inch are equivalent to 1.735 atmospheres.

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3 years ago
A physics student throws a softball straight up into the air. The ball was in the air for a total of 3.56 s before it was caught
meriva

Answer:

The initial velocity of the softball is 14.711 meters per second.

Explanation:

This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.

From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:

y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2} (Eq. 1)

Where:

y_{o} - Initial height of the softball, measured in meters.

y - Final height of the softball, measured in meters.

v_{o} - Initial velocity of the softball, measured in meters per second.

t - Time, measured in seconds.

g - Gravitational acceleration, measured in meters per square second.

If we know that y = y_{o}, t = 3.56\,s and g = -9.807\,\frac{m}{s^{2}}, the initial velocity of the softball is:

v_{o}\cdot (3\,s)+\frac{1}{2}\cdot (-9.807\,\frac{m}{s^{2}} )\cdot (3\,s)^{2} = 0

3\cdot v_{o} -44.132\,m= 0

v_{o} = 14.711\,\frac{m}{s}

The initial velocity of the softball is 14.711 meters per second.

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