Answer:
A. Part a is the attachment
B. total work = 10.4kj
Explanation:

T1 = constant temperature
nRT1 = PaVa = PbVb
We write equation as

5ma = Pa, 5L = Va, Vb = 10L(temperature is doubled)

W1 = 25 ln2
W1 = 25 x 0.693
= 17.327kj
The isochoric expansion has no change in volume. So,
W2 = 0
Isothermal compression

T3 = constant temperature
nRT3 = PcVc = PdVd

Pc = 1mpa Vc = 10L Vd = 5L

= 10x-0.693
= -6.93kj
Isochoric compression has no change in volume. Workdone w4 = 0
Total workdone = w1 + w2 + w3 + w4
= 17.33 + 0 + (-6.93) + 0
= 10.4kj
Answer:
nothing
Explanation:If you ride a bike around the block and return to the exact point where you started, your displacement is zero.
By definition, displacement involves changing an object from its original position. No matter how far or for how long a body moves, if it returns to the position it started from, it has not been displaced at all. This means that the body has zero displacements.
Answer:
v = 8.65 m/s
Explanation:
Given that,
Distance covered by the doge, d = 45 m
Time taken, t = 5.2 s
We need to find its average speed. The total distance covered divided by the total time taken is called the average speed of an object. So,

So, the average speed is 8.65 m/s.
Answer:
a) v₃ = 19.54 km, b) 70.2º north-west
Explanation:
This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition
vector 1 moves 26 km northeast
let's use trigonometry to find its components
cos 45 = x₁ / V₁
sin 45 = y₁ / V₁
x₁ = v₁ cos 45
y₁ = v₁ sin 45
x₁ = 26 cos 45
y₁ = 26 sin 45
x₁ = 18.38 km
y₁ = 18.38 km
Vector 2 moves 45 km north
y₂ = 45 km
Unknown 3 vector
x3 =?
y3 =?
Vector Resulting 70 km north of the starting point
R_y = 70 km
we make the sum on each axis
X axis
Rₓ = x₁ + x₃
x₃ = Rₓ -x₁
x₃ = 0 - 18.38
x₃ = -18.38 km
Y Axis
R_y = y₁ + y₂ + y₃
y₃ = R_y - y₁ -y₂
y₃ = 70 -18.38 - 45
y₃ = 6.62 km
the vector of the third leg of the journey is
v₃ = (-18.38 i ^ +6.62 j^ ) km
let's use the Pythagorean theorem to find the length
v₃ = √ (18.38² + 6.62²)
v₃ = 19.54 km
to find the angle let's use trigonometry
tan θ = y₃ / x₃
θ = tan⁻¹ (y₃ / x₃)
θ = tan⁻¹ (6.62 / (- 18.38))
θ = -19.8º
with respect to the x axis, if we measure this angle from the positive side of the x axis it is
θ’= 180 -19.8
θ’= 160.19º
I mean the address is
θ’’ = 90-19.8
θ = 70.2º
70.2º north-west