M1U1 + M2V2 = (M1+M2)V, where M1 is the mass of the moving car, M2 is the mass of the stationary car, U1 is the initial velocity, and V is the common velocity after collision.
therefore;
(1060× 16) + (1830 ×0) = (1060 +1830) V
16960 = 2890 V
V = 5.869 m/s
The velocity of the cars after collision will be 5.689 m/s
In 6 secs, the dog covers-
S=vt
8.9*6 = 53.4 m.
In the same time, the cat covers, 53.4-3.8 = 49.6 m.
Thus, speed of the cat, v= s/t,
= 49.6/6 = 8.267 m/s
<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: false
Explanation:
You have to make sure it doesn’t stay wet
The force is still 1 pound.
The TORQUE around the pivot is (force) x (distance from the pivot) = 1 foot-pound. (B)
