Answer:
It acts perpendicular to the velocity and is directed toward the center of the circle.
Explanation:
For a force to make an object move in a circular motion, the force must act perpendicular to the velocity and directed toward the center of the circle of motion as shown in the attached image.
Hello.
The answer is:
It creates a spark.
Then thespark can sometimes start a fire or other serious problems.
Have a nice day
<h3>Answer: 22.48°</h3><h3 /><h3>Explanation:</h3>
refractive index = sin i / sin e
where i is the angle of incidence
e is the angle of refraction
1.5 = sin 35 / sin e
1.5 = 0.5736/sin e
sin e = 0.5736/ 1.5
sin e = 0.3824
e = 22.48°
<h3 />
Answer: zero.
Justification:
The downward velocity of the sky diver just before starting to fall is zero, assuming that the helicopter is not moving but just hovering.
Before starting to fall, the velocity of the skydiver is the same of the helicopter, which is zero. It is only, once she jumps out of the helicopter that her weight is not supported by the helicopter and so the gravitational force of the Earth attracts the skydiver and she starts to gain velocity at a rate equal to g (around 9.81 m/s²).
This is a projectile motionm with
Iniital height, Yo = 2.4 m
Iniital vertical velocity, Voy = 0
Kind of motion: vertical fall => Y = Yo - Voy * t - [g*t^2] / 2
Initial horizontal position, Xo =0
Initial horizontal velocity, Vox = 22.3 m/s
kind of motion: uniform in the horizontal direction => X = Vox *t
You can find:
a) Height when the ball is passing the net, which is at 11.9 m from the service point.
Time when X = 11.9 m
X = Vox*t => t = X / Vox = 11.9 m / 22.3m/s = 0.5336 s
Replace that time into the vertical equation to find the height, Y:
Y = Yo - Voy*t - g*(t^2) / 2 = 2.4m - 0 - 9.8 m/s^2 * (0.5336 s)^2 / 2 = 1.0 m
Then, given that height of the net is 0.91 m, the ball will pass 0.09m over the net (9 cm).
b) How far from the net the ball will hit the ground, given that it is 11.9 m away of the service point.
From the vertical equation you can determine the flight time:
Y = Yo - g(t^2) / 2 => 0 = 2.4 - 4.9 t^2 => 4.9t^2 = 2.4 = > t^2 = 2.4 / 4.9 = 0.49 => t = √0.49 = 0.7 s
Now use that time in the vertical equation to find the X position of the ball when it hits the ground:
X = Vox *t = 22.3m/s * 0.7 s = 15.6 m
Then, subtract the position of the net to find how far away from it the ball hits the ground: 15.6 - 11.9 =3.7 m.