Answer:
3.6 m
Explanation:
let x = horizontal distance between emily and allison should be for allison to catch the ball
Find horizontal speed of the ball
vx = 12 sin 30 = 12 x 0.5 = 6 m/s
To find time taken, we will use vertical values of the ball motion
Initial velocity in vertical direction
u = 12 cos 30 = 10.392 m/s
let a = g = 9.8m/s2
Use equation of motion
s = ut +1/2at^2
s = vertical distance = 8
8 = (10.392)t + (1/2)(9.8)t^2
8 = (10.392)t + (4.9)t^2
4.9t^2 + 10.392t - 8 = 0
Using formula of quadratic or calculator, we'll find
t = 0.6 and t = -2.72
We pick t=0.6s since it's not logical time in negative
Assuming no air resistance or external forces, the ball will move 6m/s horizontally. Hence using the formula of speed
speed vx = distance x / time
x = (vx)(t)
= 6 x 0.6
= 3.6 m
If you drop a <span>6.0x10^-2 kg ball from height of 1.0m above hard flat surface, and a</span>fter the ball had bounce off the flat surface, the kinetic energy of the ball would be mgh - 0.14 = 0.45.
<h2><u>Answer:</u></h2>
The simulation kept track of the variables and automatically recorded data on object displacement, velocity, and momentum. If the trials were run on a real track with real gliders, using stopwatches and meter sticks for measurement, the data compared by the following statements:
1. (There would be variables that would be hard to control, leading to less reliable data.)
3. (Meter sticks may lack precision or may be read incorrectly.)
4. (Real glider data may vary since real collisions may involve loss of energy.)
5. (Human error in recording or plotting the data could be a factor.)
Answer:
the expression of current in the loop enclosed to the left of the capacitor plate is
Explanation:
As we know by Ampere's law that line integral of magnetic field around a closed loop is proportional to the current enclosed in the path
So we will have
so we have
so above is the expression of current in the loop enclosed to the left of the capacitor plate
Answer:
A) 12P
Explanation:
The power produced by a force is given by the equation
where
W is the work done by the force
T is the time in which the work is done
At the beginning in this problem, we have:
W = work done by the force
T = time taken
So the power produced is
Later, the force does six times more work, so the work done now is
And this work is done in half the time, so the new time is
Substituting into the equation of the power, we find the new power produced:
So, 12 times more power.
