A)
It is a launch oblique, therefore the initial velocity in the vertical direction is zero. Space Hourly Equation in vertical, we have:
Through Definition of Velocity, comes:
B)
Using the Velocity Hourly Equation in vertical direction, we have:
The angle of impact is given by:
If you notice any mistake in my english, please let me know, because i am not native.
Answer:
8.1km/h Northwest
Explanation:
The 8.1km/h Northwest gives the best description of her distance from start to finish. This distance can be represented in a right angle triangle , this is the hypotenuse which is the longest side of the triangle. If we add 5.7 and 5.8 this gives 11.5km/h compared to 8.1km/h which is a smaller distance and the best.
Answer:
72.54 degree west of south
Explanation:
flow = 3.9 m/s north
speed = 11 m/s
to find out
point due west from the current position
solution
we know here water is flowing north and ship must go south at an equal rate so that the velocities cancel and the ship just goes west
so it become like triangle with 3.3 point down and the hypotenuse is 11
so by triangle
hypotenuse ×cos(angle) = adjacent side
11 ×cos(angle) = 3.3
cos(angle) = 0.3
angle = 72.54 degree west of south
Answer:
Densities increase down the group
MP and BP decrease down the group
Softness increased going down the group
Speed of reacting increases going down the group
Answer:
the height reached is = 0.458 [m]
Explanation:
We need to make a sketch of the ball and see the location of the reference point where the potential energy is zero. But the kinetic energy will be defined by the following expression:
![Ek=\frac{1}{2} *m*v^{2} \\where:Ek= kinetic energy [J]\\m = mass of the ball [kg]\\v = velocity of the ball [m/s]](https://tex.z-dn.net/?f=Ek%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%5C%5Cwhere%3AEk%3D%20kinetic%20energy%20%5BJ%5D%5C%5Cm%20%3D%20mass%20of%20the%20ball%20%5Bkg%5D%5C%5Cv%20%3D%20velocity%20of%20the%20ball%20%5Bm%2Fs%5D)
Replacing the values on the equation we have:
![Ek=\frac{1}{2}*(2)*(3^{2} )\\ Ek=9[J]\\](https://tex.z-dn.net/?f=Ek%3D%5Cfrac%7B1%7D%7B2%7D%2A%282%29%2A%283%5E%7B2%7D%20%29%5C%5C%20Ek%3D9%5BJ%5D%5C%5C)
This kinetic energy will be transformed in potential energy in the moment when the ball starts to rolling up. Therefore the maximum height reached by the ball depends of the initial velocity given to the ball.
![Ek=Ep\\where\\Ep=potential energy [J]\\Ep=m*g*h\\where\\g=gravity = 9.81[m/s^2]\\h=height reached [m]\\](https://tex.z-dn.net/?f=Ek%3DEp%5C%5Cwhere%5C%5CEp%3Dpotential%20energy%20%5BJ%5D%5C%5CEp%3Dm%2Ag%2Ah%5C%5Cwhere%5C%5Cg%3Dgravity%20%3D%209.81%5Bm%2Fs%5E2%5D%5C%5Ch%3Dheight%20reached%20%5Bm%5D%5C%5C)
Now we have:
![h=\frac{Ep}{m*g} \\h=\frac{9}{2*9.81} \\\\h=0.45 [m]](https://tex.z-dn.net/?f=h%3D%5Cfrac%7BEp%7D%7Bm%2Ag%7D%20%5C%5Ch%3D%5Cfrac%7B9%7D%7B2%2A9.81%7D%20%5C%5C%5C%5Ch%3D0.45%20%5Bm%5D)
In that moment when the ball reach the 0.45 [m] the potencial energy will be maximum and equal to the kinetic energy when the ball has a velocity of 3[m/s]
