Answer:
it's CROSSOVER I knew it because I'm a table tennis player
In this problem, we use moment balance to get how far the boat moves. In this case, we multiply the mass of an object multiplied by the distance of the object. In this case, the balance goes:
1326 kg * 6.5 m = 1.2 x 104 kg * x where x is the distance
x is equal to 0.71825 meters.
Given:
heat generated by John's cooling system,
= 45 W (1)
If ρ, A, and v corresponds to John's cooling system then let
be the variables for Mike's system then:
![\rho = 9.5\rho_{1}](https://tex.z-dn.net/?f=%5Crho%20%20%3D%209.5%5Crho_%7B1%7D)
![\rho_{1} = \frac{\rho}{9.5}](https://tex.z-dn.net/?f=%5Crho_%7B1%7D%20%20%3D%20%5Cfrac%7B%5Crho%7D%7B9.5%7D)
![v_{1} =3.5 v](https://tex.z-dn.net/?f=v_%7B1%7D%20%3D3.5%20v)
Formula use:
Heat generated, ![H = \rho A v^{3}](https://tex.z-dn.net/?f=H%20%3D%20%5Crho%20A%20v%5E%7B3%7D)
where,
= density
A = area
v = velocity
Solution:
for Mike's cooling system:
=
⇒
=
× A ×
= 4.513
A ![\rho](https://tex.z-dn.net/?f=%5Crho)
Using eqn (1) in the above eqn, we get:
= 4.513 × 45 = 203.09 W
Answer:
Explanation:
Range= U²sin2theta/g
= 102²* sin (2*42)/9.8
= 1054.74m
Yes she will be abable to jump
The distance from the edge will be
(1054.74- 420)m
= 634.74m