<span>The ball clears by 11.79 meters
Let's first determine the horizontal and vertical velocities of the ball.
h = cos(50.0)*23.4 m/s = 0.642788 * 23.4 m/s = 15.04 m/s
v = sin(50.0)*23.4 m/s = 0.766044 * 23.4 m/s = 17.93 m/s
Now determine how many seconds it will take for the ball to get to the goal.
t = 36.0 m / 15.04 m/s = 2.394 s
The height the ball will be at time T is
h = vT - 1/2 A T^2
where
h = height of ball
v = initial vertical velocity
T = time
A = acceleration due to gravity
So plugging into the formula the known values
h = vT - 1/2 A T^2
h = 17.93 m/s * 2.394 s - 1/2 9.8 m/s^2 (2.394 s)^2
h = 42.92 m - 4.9 m/s^2 * 5.731 s^2
h = 42.92 m - 28.0819 m
h = 14.84 m
Since 14.84 m is well above the crossbar's height of 3.05 m, the ball clears. It clears by 14.84 - 3.05 = 11.79 m</span>
Answer:
80m/s
Explanation:
to find it you have to work it out by using the formula distance divided by speed to find time.
Answer with Explanation:
We are given that
Initial velocity,u=4.5 m/s
Time=t =0.5 s
Final velocity=v=0m/s
We have to find the deceleration and estimate the force exerted by wall on you.
We know that
Acceleration=
Using the formula
Acceleration=
deceleration=a=
We know that
Force =ma
Using the formula and suppose mass of my body=m=40 kg
The force exerted by wall on you
Force=
Answer:
Reactance
Explanation:
In an AC circuit, the capacitive reactance of a capacitor is given by:

where
f is the frequency of the AC current
C is the capacitance of the capacitor
The reactance of the capacitor tells somehow the "resistance" of the capacitor to the passage of current through it. In fact:
- When the frequency of the AC current is zero (this means, we are in regime of DC current), the reactance becomes infinite, and this is true because the capacitor does not let the current pass through it)
- When the frequency of the AC current tends to infinite, the reactance becomes zero, and this is true because in this case the current changes direction so fast that the capacitor has not enough time to "block" the current, so the current almost no feels the presence of the capacitor.
Answer: did you get the answers?
Explanation: