The second ball traveled a greater distance when compared to the first ball because the second ball spent more time in motion.
The given parameters;
- time of fall of the first ball, t = 1 s
- time of fall of the second ball, t = 3 s
The distance traveled by each ball is calculated using the second equation of motion as shown below.
The distance traveled by the first ball is calculated as follows;
The distance traveled by the second ball is calculated as follows;
Thus, the second ball traveled a greater distance because it spent more time in motion.
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Answer:
I dont know what to answer so im just gonna say ok
He gets heavier due to the upward acceleration of the lift
There are some missing data in the problem. The full text is the following:
"<span>A </span>real<span> (</span>non-Carnot<span>) </span>heat engine<span>, </span>operating between heat reservoirs<span> at </span>temperatures<span> of 710 K and 270 K </span>performs 4.1 kJ<span> of </span>net work<span>, and </span>rejects<span> 9.7 </span>kJ<span> of </span>heat<span>, in a </span>single cycle<span>. The </span>thermal efficiency<span> of a </span>Carnot heat<span> engine, operating between the same </span>heat<span> reservoirs, in percent, is closest to.."
Solution:
The efficiency of a Carnot cycle working between cold temperature </span>
and hot temperature
is given by
and it represents the maximum efficiency that can be reached by a machine operating between these temperatures. If we use the temperatures of the problem,
and
, the efficiency is
Therefore, the correct answer is D) 62 %.
Answer:
-9.4 kg m/s
Explanation:
Change in momentum = final momentum − initial momentum
Δp = p − p₀
Δp = -4.3 kg m/s − 5.1 kg m/s
Δp = -9.4 kg m/s
Impulse = change in momentum
J = Δp
J = -9.4 kg m/s
