Answer:
<em>D. remains constant throughout the fall.</em>
Explanation:
<u>Horizontal Launching
</u>
We can launch an object in free air in three forms: vertically, horizontally or inclined. In any case, the only acting force to modify the object's velocity and make it go back to the ground is the force of gravity and it's always directed downwards. Unless friction or air resistance is considered, the horizontal motion is not affected because no force is acting in that direction.
The rock described in the question was launched at 3 m/s pointed horizontally. Immediately after launching, the rock starts to fall to the ground and gain vertical velocity, but the horizontal component remains the same until it completes the flight.
The D option is correct: the horizontal velocity of the rock remains constant throughout the fall
The equation of movement for this case is given by:
Where,
vf: final speed
a: acceleration
t: time
vo: initial speed
Substituting values we have:
Clearing the acceleration we have:
Answer:
his acceleration during this time is:
![a = -0.225 \frac{m}{s ^ 2}](https://tex.z-dn.net/?f=%20a%20%3D%20-0.225%20%5Cfrac%7Bm%7D%7Bs%20%5E%202%7D%20)
Answer:
Time = 2758.62 seconds
Explanation:
Given the following data;
Speed = 290 m/s
Distance = 800 km to meters = 800 * 1000 = 800000
To find the time;
Time = distance/speed
Time = 800000/290
Time = 2758.62 seconds
Therefore, it will take the pilot 2758.62 seconds to reach the airport.
Answer:
If white light shines on a red ball, the ball reflects back mostly red light, and so we see red. Most of the greens and blues that are part of white light are absorbed by the ball so we cannot see them. Likewise, a blue book is reflecting the blue part of the white light spectrum.
Answer:
(a) the runner's kinetic energy at the given instant is 308 J
(b) the kinetic energy increased by a factor of 4.
Explanation:
Given;
mass of the runner, m = 64.1 kg
speed of the runner, u = 3.10 m/s
(a) the kinetic energy of the runner at this instant is calculated as;
![K.E_i = \frac{1}{2} mu^2\\\\K.E_i = \frac{1}{2} \times 64.1 \times 3.1^2\\\\K.E_i = 308 \ J](https://tex.z-dn.net/?f=K.E_i%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mu%5E2%5C%5C%5C%5CK.E_i%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%20%5Ctimes%2064.1%20%5Ctimes%203.1%5E2%5C%5C%5C%5CK.E_i%20%3D%20308%20%5C%20J)
(b) when the runner doubles his speed, his final kinetic energy is calculated as;
![K.E_f = \frac{1}{2} mu_f^2\\\\K.E_f = \frac{1}{2} m(2u)^2\\\\K.E_f = \frac{1}{2} \times 64.1 \ \times (2\times 3.1)^2\\\\K.E_f = 1232 \ J](https://tex.z-dn.net/?f=K.E_f%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mu_f%5E2%5C%5C%5C%5CK.E_f%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20m%282u%29%5E2%5C%5C%5C%5CK.E_f%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%2064.1%20%5C%20%5Ctimes%20%282%5Ctimes%203.1%29%5E2%5C%5C%5C%5CK.E_f%20%3D%201232%20%5C%20J)
the change in the kinetic energy is calculated as;
![\frac{K.E_f}{K.E_i} = \frac{1232}{308} =4](https://tex.z-dn.net/?f=%5Cfrac%7BK.E_f%7D%7BK.E_i%7D%20%3D%20%5Cfrac%7B1232%7D%7B308%7D%20%3D4)
Thus, the kinetic energy increased by a factor of 4.