Answer:
39.7 m
Explanation:
First, we conside only the last second of fall of the body. We can apply the following suvat equation:
where, taking downward as positive direction:
s = 23 m is the displacement of the body
t = 1 s is the time interval considered
is the acceleration
u is the velocity of the body at the beginning of that second
Solving for u, we find:
Now we can call this velocity that we found v,
v = 18 m/s
And we can now consider the first part of the fall, where we can apply the following suvat equation:
where
v = 18 m/s
u = 0 (the body falls from rest)
s' is the displacement of the body before the last second
Solving for s',
Therefore, the total heigth of the building is the sum of s and s':
h = s + s' = 23 m + 16.7 m = 39.7 m
Answer:
a)
The direction will be negative direction.
b)
The direction will be positive direction.
Explanation:
Given that
q1 = +7.7 µC is at x1 = +3.1 cm
q2 = -19 µC is at x2 = +8.9 cm
We know that electric filed due to a charge given as
Now by putting the va;ues
a)
The net electric field
The direction will be negative direction.
As we know that electric filed line emerge from positive charge and concentrated at negative charge.
b)
Now
distance for charge 1 will become =5.5 - 3.1 = 2.4 cm
distance for charge 2 will become =8.9-5.5 = 3.4 cm
The net electric field
The direction will be positive direction.
Answer:
At time 10.28 s after A is fired bullet B passes A.
Passing of B occurs at 4108.31 height.
Explanation:
Let h be the height at which this occurs and t be the time after second bullet fires.
Distance traveled by first bullet can be calculated using equation of motion
Here s = h,u = 450m/s a = -g and t = t+3
Substituting
Distance traveled by second bullet
Here s = h,u = 600m/s a = -g and t = t
Substituting
Solving both equations
So at time 10.28 s after A is fired bullet B passes A.
Height at t = 7.28 s
Passing of B occurs at 4108.31 height.
