Answer:
10 days
Explanation:
The half-life of a radioactive sample is the time taken for half of the sample to decay.
In the diagram, the half-life corresponds to the time after which the % of cobalt-57 has halved. We can observe the following:
At t=10 days, the % of Co remaining is approximately 45%
At t=20 days, the % of Co remaining is approximately 22%
This means that the sample of cobalt-57 has halved in 10 days, so the half-life of cobalt-57 is 10 days.
The work done by the heat engine is 40 kCal.
The given parameters;
- input heat of the engine, Q₁ = 70 kCal
- output heat of the engine, Q₂ = 30 kCal
To find:
- the work done by the heat engine
The work done by the heat engine is the change in the heat energy of the engine;
W = Q₂ - Q₁
Substitute the given parameters and solve work done (W)
W = 70 kCal - 30 kal
W = 40 kCal
Thus, the work done by the heat engine is 40 kCal.
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So u have to use The gravitational law: F = (GmM)/(r^2) because for the first one they are the same distance apart but both masses are twice as big the bottom force is 4 times larger
But for the second one all masses are the same but the bottom is twice as distant so the top is 4 times larger