= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)
= -0.00225 m
New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248
D
PLEASE HELPP MEEE!!!!!!!!!The goal is to increase the power; therefore, it is necessary to
2 answers:
As we know that power is defined as rate of work done
so we will have
so in order to increase the power as per above formula we know that either we need to increase the work or we need to decrease the time to complete that work
So here the correct answer will be
increase the work being done or decrease the time in which the work is completed.
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= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)
= -0.00225 m
New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248
D
The distance between Mars and the Sun in the scale model would be 1140 m
Explanation:
In this scale model, we have:
represents an actual distance of
The actual distance between Mars and the Sun is 228 million km, therefore
On the scale model, this would corresponds to a distance of .
Therefore, we can write the following proportion:
And solving for , we find:
Learn more about distance:
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