I would expect it to be slightly basic.
Answer:
same 0.81m
Explanation:
in this problem if we assume there no resistance of any sort. and we apply the energy conservation
change in Potential energy = change in kinetic energy
mgh = 0.5mv^2
gh = 0.5v^2
the above relation suggests that the speed at the bottom is only depending on the height it is released from not on the shape, mass or radius.
so at the bottom
put h = 0.81m
9.81 * 0.81 * 2 = v^2
v=3.99 m/s
both CYLINDER and SPHERE will have same velocity at the bottom if released from the same height irrespective of shape and size
Answer:
<h2>E) 52.5 cm</h2>
Explanation:
Step one:
given data
period T= 3 milliseconds= 0.003
velocity v= 175m/s
wave lenght λ=?
Step two:
we know that f=1/T
the expression relating period and wave lenght is
v=λ/T
λ=v*T
λ=175*0.002
λ=0.525m
to cm= 0.525*100
=52.5cm
The wavelength of the wave is E) 52.5 cm
Answer:
We kindly invite you to read carefully the explanation and check the image attached below.
Explanation:
According to this problem, the rocket is accelerated uniformly due to thrust during 30 seconds and after that is decelerated due to gravity. The velocity as function of initial velocity, acceleration and time is:
(1)
Where:
- Initial velocity, measured in meters per second.
- Final velocity, measured in meters per second.
- Acceleration, measured in meters per square second.
- Initial time, measured in seconds.
- Final time, measured in seconds.
Now we obtain the kinematic equations for thrust and free fall stages:
Thrust (
, 
, 
, 
)
(2)
Free fall (
, 
, 
, 
)
(3)
Now we created the graph speed-time, which can be seen below.
D ............................
