Answer:
-4.72005 m/s
0 m/s
Explanation:
Displacement = 5220 km
Time = 12.8 days
The average velocity is given by
The average for the return flight is -4.72005 m/s
When the total displacement is divided by the total time of a journey then we get the average velocity.
Displacement is the minimum distance between the initial and final points of the journey.
Here, the displacement of the whole episode is 0 as the initial and final point is zero.
Hence, the average velocity for the whole episode is 0 m/s
Answer:
vertical force cannot change the velocity on the x-axis. t =x/v₀ₓ
Explanation:
The force is a vector magnitude, so the forces on the x-axis affect the acceleration on this axis. Consequently a vertical force cannot change the velocity on the x-axis.
= m g
Fₓ = 0
The horizontal velocity in projectile motion is constant, if we neglect the air resistance, so it can be used to find the time of a horizontal displacement
x = v₀ₓ t
t =x/v₀ₓ
The only magnitude that is the same for both movements is the time that is a scalar
Answer:
Ek1 = 900000 [J]
Ek1 = 400000 [J]
Explanation:
In order to solve this problem we must remember that kinetic energy is defined as the product of mass by velocity squared by a medium. Therefore using the following equation we have:
where:
m = mass = 500 [kg]
v1 = 60 [m/s]
So we have:
Ek1 = 0.5*500*(60^2)
Ek1 = 900000 [J]
and:
Ek2 = 0.5*500*(40^2)
Ek2 = 400000 [J]
Answer:
The variable manipulated or controlled by the experimenter is called the independent variable.
Example:
If the flow velocity at the bottom of a tank is measured by varying the height of water in the tank, we are measuring velocity as a function of water height.
Therefore,
water height = independent variable (controlled)
velocity = dependent variable (measured in response to water height).
Mathematically,
v = f(h)
where v = response variable (dependent)
h = controlled variable (independent).
Answer:
The kinetic energy is: 50[J]
Explanation:
The ball is having a potential energy of 100 [J], therefore
PE = [J]
The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.
![E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3Dm%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cg%3D%20gravity%5Bm%2Fs%5E%7B2%7D%20%5D%5C%5Cm%20%3D%20mass%20%5Bkg%5D%5C%5Cm%3D%20%5Cfrac%7BE_%7Bp%7D%20%7D%7Bg%2Ah%7D%5C%5C%20m%3D%20%5Cfrac%7B100%7D%7B9.81%2A10%7D%5C%5C%5C%5Cm%3D%201.01%5Bkg%5D%5C%5C%5C%5C)
In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.
When the elevation is 5 [m], we have a potential energy of
![P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\](https://tex.z-dn.net/?f=P_%7Be%7D%20%3Dm%2Ag%2Ah%5C%5CP_%7Be%7D%20%3D1.01%2A9.81%2A5%5C%5C%5C%5CP_%7Be%7D%20%3D%2050%20%5BJ%5D%5C%5C)
This energy is equal to the kinetic energy, therefore
Ke= 50 [J]
