Answer:
The value is
Explanation:
From the we are told that
The initial speed of the object is
The greatest height it reached is 
Generally from kinematic equation we have that

At maximum height v = 0 m/s
So

=> 
Here H is the height from the initial height to the maximum height
So the initial height is mathematically represented as

=> 
=> 
Generally the time taken for the object to reach maximum height is mathematically evaluated using kinematic equation as follows

At maximum height v = 0 m/s

=> 
Generally the time taken for the object to move from the maximum height to the ground is mathematically using kinematic equation as follows

Here the initial velocity is 0 m/s given that its the velocity at maximum height
Also g is positive because we are moving in the direction of gravity
So

=> 
Generally the total time taken is mathematically represented as

=> 
=>
Answer:
0.4113772 s
Explanation:
Given the following :
Mass of bullet (m1) = 8g = 0.008kg
Initial horizontal Velocity (u1) = 280m/s
Mass of block (m2) = 0.992kg
Maxumum distance (x) = 15cm = 0.15m
Recall;
Period (T) = 2π√(m/k)
According to the law of conservation of momentum : (inelastic Collison)
m1 * u1 = (m1 + m2) * v
Where v is the final Velocity of the colliding bodies
0.008 * 280 = (0.008 + 0.992) * v
2.24 = 1 * v
v = 2.24m/s
K. E = P. E
K. E = 0.5mv^2
P.E = 0.5kx^2
0.5(0.992 + 0.008)*2.24^2 = 0.5*k*(0.15)^2
0.5*1*5.0176 = 0.5*k*0.0225
2.5088 = 0.01125k
k = 2.5088 / 0.01125
k = 223.00444 N/m
Therefore,
Period (T) = 2π√(m/k)
T = 2π√(0.992+0.008) / 233.0444
T = 2π√0.0042910
T = 2π * 0.0655059
T = 0.4113772 s
The distance quantity/ measurement must be squared.
Answer:

Explanation:
= Permittivity of free space = 
A = Area
h = Altitude = 600 m
Electric flux through the top would be
(negative as the electric field is going into the volume)
At the bottom

Total flux through the volume

Electric flux is given by

Charge per volume is given by

The volume charge density is 
Answer:
70.07 Hz
Explanation:
Since the sound is moving away from the observer then
and
when moving towards observer
With
of 76 then taking speed in air as 343 m/s we have


Similarly, with
of 65 we have

Now

v_s=27.76 m/s
Substituting the above into any of the first two equations then we obtain
