Answer:
Explanation:
Given
mass of core
Average specific heat 
And rate of increase of temperature =
Now
P=
Thus ![\frac{\mathrm{d}T}{\mathrm{d} t}=[tex]\frac{1.60\times 10^5\times 0.3349}{150\times 10^6}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%7DT%7D%7B%5Cmathrm%7Bd%7D%20t%7D%3D%5Btex%5D%5Cfrac%7B1.60%5Ctimes%2010%5E5%5Ctimes%200.3349%7D%7B150%5Ctimes%2010%5E6%7D)
A. To find work we need to know F and S; to find power we need to know F and V
I think it’s B hope it helps:)
Answer:
1.08 s
Explanation:
From the question given above, the following data were obtained:
Height (h) reached = 1.45 m
Time of flight (T) =?
Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:
Height (h) = 1.45 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
1.45 = ½ × 9.8 × t²
1.45 = 4.9 × t²
Divide both side by 4.9
t² = 1.45/4.9
Take the square root of both side
t = √(1.45/4.9)
t = 0.54 s
Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).
Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:
Time (t) taken to reach the height = 0.54 s
Time of flight (T) =?
T = 2t
T = 2 × 0.54
T = 1.08 s
Therefore, it will take the kangaroo 1.08 s to return to the earth.
