This question was apprently selected from the "Sneaky Questions" category.
The store is 3 km from his home, and he walks there with a speed of 6 km/hr. So it takes him (3 km) / (6 km/hr) = 1/2 hour to get to the store.
That's 30 minutes. So the whole part-(a.) of the question refers to only that part of the trip, and we don't care what happens once he reaches the store.
a). Over the first 30 minutes of his travel, Greg walks 3.0 km on a straight road, and he ends up 3.0 km away from where he started.
Average speed = (distance/time) = (3.0 km) / (1/2 hour) = <em>6.0 km/hr</em>
Average velocity = (displacement/time) = (3.0 km) / (1/2 hour) = <em>6.0 km/hr</em>
There's probably some more questions in part-(b.) where you'd need to use Greg's return trip to find the answers, but johnaddy210 is only asking us for part-(a.).
Answer:
Dₓ = -155 sin 23° i + 0 j
Explanation:
The diagram showing the vector has been attached to this response.
As shown in the diagram,
The vector D has an x-component (also called horizontal component) of -D sinθ i. i.e
Dₓ = -D sin θ i [The negative sign shows that D lies in the negative x direction]
Where;
D = magnitude of D = 155m
θ = direction of D = 23°
Therefore;
Dₓ = -155 sin 23° i
Since Dₓ represents the x component, its unit vector, j component has a value of 0.
Therefore, Dₓ can be written in terms of D, θ and the unit vectors i and j as follows;
Dₓ = -155 sin 23° i + 0 j
Answer: Resistance = 
The approximate diameter of a penny is, <em>d</em> = 20 mm
thickness of penny is, <em>L = </em> 1.5×
mm
The area of penny along circular face is,
= 3.14×
m²
The resistivity of copper is <em>ρ</em> = 1.72 x 10-8 Ωm.
Resistance,

Answer:
0.60 m/s
Explanation:
The average velocity from t = a to t = b is:
v_avg = (x(b) − x(a)) / (b − a)
Given that x(t) = 0.36t² − 1.20t, and the time is from 1.0 to 4.0:
v_avg = (x(4.0) − x(1.0)) / (4.0 − 1.0)
v_avg = [(0.36(4.0)² − 1.20(4.0)) − (0.36(1.0)² − 1.20(1.0))] / 3.0
v_avg = [(5.76 − 4.8) − (0.36 − 1.20)] / 3.0
v_avg = [0.96 − (-0.84)] / 3.0
v_avg = 0.60
The average speed is 0.60 m/s.
Answer:
Given that
V= 0.06 m³
Cv= 2.5 R= 5/2 R
T₁=500 K
P₁=1 bar
Heat addition = 15000 J
We know that heat addition at constant volume process ( rigid vessel ) given as
Q = n Cv ΔT
We know that
P V = n R T
n=PV/RT
n= (100 x 0.06)(500 x 8.314)
n=1.443 mol
So
Q = n Cv ΔT
15000 = 1.433 x 2.5 x 8.314 ( T₂-500)
T₂=1000.12 K
We know that at constant volume process
P₂/P₁=T₂/T₁
P₂/1 = 1000.21/500
P₂= 2 bar
Entropy change given as

Cp-Cv= R
Cp=7/2 R
Now by putting the values


a)ΔS= 20.79 J/K
b)
If the process is adiabatic it means that heat transfer is zero.
So
ΔS= 20.79 J/K
We know that

Process is adiabatic



