From Carnot's theorem, for any engine working between these two temperatures:
efficiency <= (1-tc/th) * 100
Given: tc = 300k (from question assuming it is not 5300 as it seems)
For a, th = 900k, efficiency = (1-300/900) = 70%
For b, th = 500k, efficiency = (1-300/500) = 40%
For c, th = 375k, efficiency = (1-300/375) = 20%
Hence in case of a and b, efficiency claimed is lesser than efficiency calculated, which is valid case and in case of c, however efficiency claimed is greater which is invalid.
Answer:
See the answers below
Explanation:
To solve this problem we must use the following equation of kinematics.
where:
Vf = final velocity [m/s]
Vo = initial velocity [m/s]
a = acceleration [m/s²]
t = time [s]
<u>First case</u>
Vf = 6 [m/s]
Vo = 2 [m/s]
t = 2 [s]
![6=2+a*2\\4=2*a\\a=2[m/s^{2} ]](https://tex.z-dn.net/?f=6%3D2%2Ba%2A2%5C%5C4%3D2%2Aa%5C%5Ca%3D2%5Bm%2Fs%5E%7B2%7D%20%5D)
<u>Second case</u>
Vf = 25 [m/s]
Vo = 5 [m/s]
a = 2 [m/s²]
![25=5+2*t\\t = 10 [s]](https://tex.z-dn.net/?f=25%3D5%2B2%2At%5C%5Ct%20%3D%2010%20%5Bs%5D)
<u>Third case</u>
Vo =4 [m/s]
a = 10 [m/s²]
t = 2 [s]
![v_{f}=4+10*2\\v_{f}=24 [m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3D4%2B10%2A2%5C%5Cv_%7Bf%7D%3D24%20%5Bm%2Fs%5D)
<u>Fourth Case</u>
Vf = final velocity [m/s]
Vo = initial velocity [m/s]
a = acceleration [m/s²]
t = time [s]
![v_{f}=5+8*10\\v_{f}=85 [m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3D5%2B8%2A10%5C%5Cv_%7Bf%7D%3D85%20%5Bm%2Fs%5D)
<u>Fifth case</u>
Vf = final velocity [m/s]
Vo = initial velocity [m/s]
a = acceleration [m/s²]
t = time [s]
Explanation:
Take north to be positive and south to be negative.
a = (v − v₀) / t
a = (-4.5 m/s − 4.5 m/s) / 8 s
a = -1.125 m/s²
The acceleration is 1.125 m/s² south.
The sun is facing a certain side of the earth
Answer:
t = Δa / v
Explanation:
To know which option is not true, we shall fine a relationship between acceleration (a), velocity (v), time (t) and radius (r). This is illustrated below:
Acceleration can simply be defined as the rate of change of velocity with time. Mathematically, it is expressed as shown below:
Acceleration = change in velocity / time
a = Δv / t ..... (1)
But
Δv = v₂ – v₁
Substitute the value of Δv into equation (1)
a = Δv / t
a = v₂ – v₁ / t ....... (2)
From equation (1), make Δv the subject of the equation.
a = Δv / t
Cross multiply
Δv = at .... (3)
From equation (1), make t the subject of the equation.
a = Δv / t
Cross multiply
at = Δv
Divide both side by a
t = Δv /a ...... (4)
From circular motion, centripetal's force is given by:
F = mv²/r
F = ma꜀
Therefore,
ma꜀ = mv²/r
Cancel out m
a꜀ = v²/r
SUMMARY:
a = Δv / t
a = v₂ – v₁ / t
Δv = at
t = Δv /a
a꜀ = v²/r
Considering the options given in question above, t = Δa / v is not a true statement.
