The current will decrease as the resistance has now increased, meaning less current will be 'let through' the resistor. (assuming it's in series, there's no image)
Answer:
<u>Option "C":</u> "4.5 g"
Explanation:
N0 = 36 g, Let half-life is T.
t = 3 T, n is number of half lives = t / T = 3
<u>By using the decay law of radioactivity</u>
N / N0 = (1 / 2)^n
where
"N0" be the "initial amount"
"N" be the "amount left"
"n" be the "number of half-lives"
N / 36 = (1/2)^3
N / 36 = 1 / 8
N = 36 / 8 = 4.5 g
#1). Anthony does the same amount of work as Angel, with <em>more power</em>.
#2). Power = (Work)/(Time) = 41,000 J / 500 s = <em>82 watts .</em>
#3). Power = (Work) / (Time) = 83 J / 3 sec = <em>27.7 watts</em>
Answer:
Distance - 1000m
Time - 20min
Speed - ?
Use the formula of distance ÷ time = speed.
s = d/t
s = 1000m/20min
s = 50 m/min
Hope this helps, thank you !!
Answer:
It's impossible for an ideal heat engine to have non-zero power.
Explanation:
Option A is incomplete and so it's possible.
Option B is possible
Option D is related to the first lae and has nothing to do with the second law.
Hence, the correct option is C.
The ideal engine follows a reversible cycle albeit an infinitely slow one. If the work is being done at this infinitely slow rate, the power of such an engine is zero.
We can also stat the second law of thermodynamics in this manner;
It is impossible to construct a cyclical heat engine whose sole effect is the continuous transfer of heat energy from a colder object to a hotter one.
This statement is known as second form or Clausius statement of the second law.
Thus, it is possible to construct a machine in which a heat flow from a colder to a hotter object is accompanied by another process, such as work input.
