S = ut + 0.5at^2
<span>10 = 0 + 0.5(9.81)t^2 {and if g = 10 then t^2 = 2 so t ~1.414} </span>
<span>t^2 ~ 2.04 </span>
<span>t ~ 1.43 seconds</span>
<u>Given data</u>
Source temperature (T₁) = 177°C = 177+273 = 450 K
Sink temperature (T₂) = 27°C = 27+273 = 300 K
Energy input (Q₁) = 3600 J ,
Work done = ?
We know that, efficiency (η) = Net work done ÷ Heat supplied
η = W ÷ Q₁
W = η × Q₁
First determine the efficiency ( η ) = ?
Also, we know that ( η ) = (T₁ - T₂) ÷ (T₁)
= 33.3% = 0.333
Now, Work done is W = η × Q₁
= 0.33 × 3600
<em> W = 1188 J</em>
<em>Work done by the engine is 1188 J</em>
Answer:
one-third of its weight on Earth's surface
Explanation:
Weight of an object is = W = m*g
Gravity on Earth = g₁ = 9.8 m/s
Gravity on Mars = g₂ = 
g₁
Weight of probe on earth = w₁ = m * g₁
Weight of probe on Mars = w₂ = m * g₂ -------- ( 1 )
As g₂ = g₁/3 --------- ( 2 )
Put equation (2) in equation (1)
so
Weight of probe on Mars = w₂ = m * g₁ /3
Weight of probe on Mars = m * g₁ = w₁
⇒Weight of probe on Mars = Weight of probe on earth
Answer:
A controlled variable does not change during a experiment
Explanation:
it's c
