= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)
= -0.00225 m
New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248
D
Read each scenario below. Then select the answer choices that complete the sentences,
2 answers:
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= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)
= -0.00225 m
New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248
D
Answer:
a)
b)
Explanation:
a) Let's use the constant velocity equation:
- v is the speed of the muon. 0.9*c
- c is the speed of light 3*10⁸ m/s
b) Here we need to use Lorentz factor because the speed of the muon is relativistic. Hence the time in the rest frame is the product of the Lorentz factor times the time in the inertial frame.
v is the speed of muon (0.9c)
Therefore the time in the rest frame will be:
No we use the value of Δt calculated in a)
I hope it helps you!
Answer:
The temperature of the steam during the heat rejection process is 42.5°C
Explanation:
Given the data in the question;
the maximum temperature T in the cycle is twice the minimum absolute temperature T
in the cycle
T = 0.5T
now, we find the efficiency of the Carnot cycle engine
η = 1 - T
/T
η = 1 - T
/0.5T
η = 0.5
the efficiency of the Carnot heat engine can be expressed as;
η = 1 - W
/Q
where W is net work done, Q
is is the heat supplied
we substitute
0.5 = 60 / Q
Q = 60 / 0.5
Q = 120 kJ
Now, we apply the first law of thermodynamics to the system
W = Q
- Q
60 = 120 - Q
Q = 60 kJ
now, the amount of heat rejection per kg of steam is;
q = Q
/m
we substitute
q = 60/0.025
q = 2400 kJ/kg
which means for 1 kilogram of conversion of saturated vapor to saturated liquid , it takes 2400 kJ/kg of heat ( enthalpy of vaporization)
q = h
= 2400 kJ/kg
now, at h = 2400 kJ/kg from saturated water tables;
T = 40 + ( 45 - 40 ) (
)
T = 40 + (5) × (0.5)
T = 40 + 2.5
T = 42.5°C
Therefore, The temperature of the steam during the heat rejection process is 42.5°C
Answer:
v₁ = 0.375 m / s , x = 0.335 m
Explanation:
Let's analyze this interesting exercise, the block moves and has a friction force with the tile, we assume that the speed of the block is constant, so the friction force opposes the block movement. For the only force that acts (action and reaction) this friction force exerted by the block that is in the direction of movement of the tile.
We can also see that the isolated system formed by the block and the tile will reach a stable speed where friction cannot give the system more energy, this speed can be found by treating the system with the conservation of linear momentum.
initial moment. Right at the start of the movement
p₀ = m v₀ + 0
final moment. Just when it comes to equilibrium
= (m + M) v₁
how the forces are internal
p₀ =p_{f}
m v₀ = (m + M) v₁
v₁ = m /m+M v₀
let's calculate
v₁ = 0.4 /(0.4 + 2.8) 3
v₁ = 0.375 m / s
Let's apply Newton's second law to the Block, to find the friction force
Y axis
N - W = 0
N = W
N = m g
where m is the mass of the block
the friction force has the formula
fr = μ N
fr = μ m g
We apply Newton's second law to slab
X axis
fr = M a
where M is the mass of the slab
μ m g = M a
a = μ g m / M
let's calculate
a = 0.15 9.8 0.4 / 2.8
a = 0.21 m / s²
With kinematics we can find the position
v²= v₀²+2 a x
as the slab is initially at rest, its initial velocity is zero
v² = 2 a x
x = v2 / 2a
let's calculate
x = 0.375²/2 0.21
x = 0.335 m