Answer:
2.4 m/s
Explanation:
Given:
Velocity of the object moving north = 2.1 m/s
Velocity of the river moving eastward = 1.2 m/s
The resultant velocity is the vector sum of the velocities of object and river.
Since the directions of velocity of object and river are perpendicular to each other, the magnitude of the resultant velocity is obtained using Pythagoras Theorem.
The velocities are the legs of the right angled triangle and the resultant velocity is the hypotenuse.
The magnitude of the resultant velocity (R) is given as:
![R^2=2.1^2+1.2^2\\\\R^2=4.41+1.44\\\\R=\sqrt{5.85}\\\\R=2.4\ m/s](https://tex.z-dn.net/?f=R%5E2%3D2.1%5E2%2B1.2%5E2%5C%5C%5C%5CR%5E2%3D4.41%2B1.44%5C%5C%5C%5CR%3D%5Csqrt%7B5.85%7D%5C%5C%5C%5CR%3D2.4%5C%20m%2Fs)
Therefore, the resultant velocity has a magnitude of 2.4 m/s.
Basically it is the difference in velocity divided by the time it takes to make that change.
Answer:
Pnitrogen=3.18 kPa, Poxygen=1.62 kPa , Ptotal= 4.80 kPa
Explanation:
partial pressure equation becomes Ptotal = Pnitrogen + Poxygen
Partial pressure of Nitrogen
Pnitrogen= nRT/V
n=no of moles =mass/molar mass
mass of nitrogen=0.6kg
Molar mass of nitrogen gas=28gmol^-1
n=0.6/28=0.0214moles
R=0.2968 kPa·m3/kg·K
T=300k
V=0.6m^3
Pnitrogen=(0.0214 * 0.2968 * 300)/0.6
Pnitrogen=3.18 kPa
Likewise
Poxygen=nRT/V
n=0.4/32=0.0125moles
R=0.2598 kPa·m3/kg·K
T=300k
V=0.6m^3
Poxygen=(0.0125 * 0.2598 * 300)/0.6
Poxygen=1.62 kPa
Ptotal= 3.18+1.62= 4.80 kPa
Answer:
Velocity: The rate of change of displacement of an object (displacement over elapsed time) is velocity. Velocity is a vector since it has both magnitude (called speed) and direction. For example, if you drive 10 miles North in 0.25 hours (15 minutes), your velocity is 10 miles/0.25 hours = 40 mph in the northerly direction. For further explanation of vectors click here.
Acceleration: The rate of change of velocity is acceleration. Like velocity, acceleration is a vector and has both magnitude and direction. For example, a car in straight-line motion is said to have forward (positive) acceleration if it is speeding up and rearward (negative) acceleration if it is slowing down. We sometimes refer to negative acceleration as deceleration.