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.
One that can help you is:
ΔT=<span>T<span>Final</span></span>−<span>T<span>Initia<span>l
That is of course adding both tmepratures. There is one more that is a lil bit more complex
</span></span></span><span><span>Tf</span>=<span>Ti</span>−Δ<span>H<span>rxn</span></span>∗<span>n<span>rxn</span></span>/(<span>C<span>p,water</span></span>∗<span>m<span>water</span></span>)
This one is taking into account that yu can find temperature and that there could be a change with a chemical reaction. Hope this helps</span>
Answer:
101.54m/h
Explanation:
Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;
Let l be the be the distance further away at which they will meet from the current points;
#The speed toward each other.
Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h
