First, you find the velocity at each component. The general equation is:
a = (v2 - v1)/t
a,x = (v2,x - v1,x)/t
-0.105 = (v2,x - 8.57)/6.67
v2,x = 7.87 m/s
a,y = (v2,y - v1,y)/t
0.101 = (v2,y - -2.61)/6.67
v2,y = -1.94 m/s
To find the final speed, find the resultant velocity by taking the hypotenuse.
v^2 = (v2,x)^2 + (v2,y)^2
v^2 = (7.87)^2 + (-1.94)^2
v = 8.1 m/s
True ..............................
Answer:
Option B
Change in entropy of the process is 
Explanation:
The entropy of a system is a measure of the degree of disorderliness of the system.
The entropy of a system moving from process 1 to 2 is given as
recall from first law, 
hence we have, 
since the process is isothermal, 
this gives us 
integrating within the limits of 1 and 2, will give us
also from ideal gas laws,
hence we have 
This makes the correct option B
If the runner is running in a circular track then yes when something or someone is moving in a circular motion at a constant speed they are indeed accelerating. They’re accelerating because the direction of the velocity vector is changing
Answer:
T'=70.92°C
Explanation:
Given that
V= 100 L=0.1 m³
P=400 KPa
T=25°C
Work done on the air = 15 KJ
W= -15 KJ
If we assume that air is ideal gas
P V = m R T
R=0.287 KJ/kg.K for air
T= 273 + 25 = 298 K
By putting the values
P V = m R T
400 x 0.1 = m x 0.287 x 298
m=0.46 kg
From first law of thermodynamics
Q= ΔU +W
Insulated piston–cylinder , Q=0
ΔU = m Cv ΔT
ΔU = - W
Cv = 0.71 KJ/kg.k for air
0.46 x 0.71 x (T' -25) = 15
T'=70.92 °C
So the final temperature of air is T'=70.92 °C
