Explanation:
It is given that,
Inductance, 
RMS value of voltage, 
Frequency, f = 60 Hz
We need to find the energy stored at t = (1 /185) s. It is assumed that energy stored in the inductor is zero at t = 0. So,
The current flowing through the inductor is given by :
I = 0.091 A
Energy stored in the inductor is, 
U = 0.000165 Joules
Hence, this is the required solution.
You will use the height of the bridge from the ground.
Solution:
Formula to be used is y=Viy(t)+g(t^2)/2
Where:
Vi=initial velocity which is 0 m/s
y=10 m
Gravitational acceleration or g =9.8m/s^2
T= time you need
Substitute all the given to the formula
10m=(0m/s)(t)+(9.8m/s^2)(t^2)/2
10mx2=9.8m/s^2(t^2)
Now isolate the variable you want to find which is T or time
10mx2/9.8m/s^2=t^2
20m/9.8m/s^2=t^2
Square root of 2.04= square root of t^2
T=1.43 secs
The answer is 1.43 seconds
Answer:
A thermos bottle works well because:
A) Its glass walls are thin
Answer: Both cannonballs will hit the ground at the same time.
Explanation:
Suppose that a given object is on the air. The only force acting on the object (if we ignore air friction and such) will be the gravitational force.
then the acceleration equation is only on the vertical axis, and can be written as:
a(t) = -(9.8 m/s^2)
Now, to get the vertical velocity equation, we need to integrate over time.
v(t) = -(9.8 m/s^2)*t + v0
Where v0 is the initial velocity of the object in the vertical axis.
if the object is dropped (or it only has initial velocity on the horizontal axis) then v0 = 0m/s
and:
v(t) = -(9.8 m/s^2)*t
Now, if two objects are initially at the same height (both cannonballs start 1 m above the ground)
And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.
You can notice that the fact that one ball is fired horizontally and the other is only dropped does not affect this, because we only analyze the vertical problem, not the horizontal one. (This is something useful to remember, we can separate the vertical and horizontal movement in these type of problems)
