How Do We Avoid Fratenities?

How do we avoid fratenities?

 

By choosing a good friend.


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Jose Buys A New Moped. He Travels 3km South And Then 4km East. How Far Does He Need To Go To Get Back To Where He Started?

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Jose buys a new moped. He travels 3km south and then 4km east. How far does he need to go to get back to where he started?   To get back to its original position, it can either be in an opposite direction of the total distance travelled  (much longer) or opposite direction of the displacement (shorter than the total distance travelled). For the total distance travelled: 3 km + 4 km = 7 km Thus, Jose needs to go back 7 km in an opposite direction (4 km West and 3 km North). For the displacement: Let d represent as displacement. Let R represent as resultant. Let Φ represents as the angle. Given: d1 = 3 km, 270° d2 = 4 km, 0° Formula: For horizontal x-axis: (n)(cosΦ) For vertical y-axis: (n)(sinΦ) For the resultant (or the magnitude displacement): R = √((x)^2 + (y)^2) For the angle (althought unnecessary): Φ = arctan (y/x) Equation: d1x = (3 km)(cos270) = 0 km d1y = (3 km)(sin270) = -3 km d2x = (4 km)(cos0) = 4 km d2y = (4 km)(sin0) = 0 km Summation of dx = (d1 + d2)x = 4 km Summatio...
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Below Are The Applications Of Electromagnetic Waves . State The Type Of Electromagnetic Wave Used In Each Application. 1. Satellite 2. Texting 3. T.V

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:Read Each Item Carefully And Supply The Required Imformation., A Bicycle Tire Was Inflated To A Pressure Of 3.74 Atm During Early Morning When The Te

:read each item carefully and supply the required imformation. A bicycle tire was inflated to a pressure of 3.74 atm during early morning when the temperature was 15°C.At noontime,the temperature rose to 35° C what was the resulting pressure in the tire (assuming that its volume did not change)?show your computation..   Good Day Problem: A bicycle tire was inflated to a pressure of 3.74 atm during early morning when the temperature was 15°C.At noontime,the temperature rose to 35° C what was the resulting pressure in the tire (assuming that its volume did not change)?show your computation Note: Kelvin = °C  + 273.15 K = 15 + 273.15 = 288.15 K K = 35 + 273.15  = 308.15 K Given Data: P₁ (initial pressure) = 3.74 atm T₁ (initial temperature) =  15°C or 288.15 K P₂ (final pressure)  =  ? unknown T₂  (final temperature) = 35°C  or 308.15 K Solution: Gay-Lussacs Law states that pressure is directly proportional to temperature when volume and amount of gas are constant. Formula = P₁/T₁...
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