Pascal's law of fluid transfer states that when there is an increase in fluid pressure, the rest of the extrinsic variables also increases. For example, in a flow of liquid in an orifice, there is a contraction of diameter in the orifice part. The fluid that will go in there increases in pressure and thereby an increase in velocity as well.
Answer:
The height of the building is 8,302.5 m
Explanation:
Given;
velocity of the projectile, u = 36 m/s
time of motion, t = 45 s
Let the upward direction of the bullet be negative,
The height of the building is calculated as;

Answer:
T = 0.003 s
(Period is written as T)
Explanation:
Period = time it takes for one wave to pass (measured in seconds)
frequency = number of cycles that occur in 1 second
(measured in Hz / hertz / 1 second)
Period : T
frequency : f
So, if we know that the frequency of a wave is 300 Hz, we can find the period of the wave from the relation between frequency and period
T =
f = 
to find the period (T) of this wave, we need to plug in the frequency (f) of 300
T = 
T = 0.00333333333
So, the period of a wave that has a frequency of 300 Hz is 0.003 s
[the period/T of this wave is 0.003 s]
The centripetal acceleration is responsible
for the artificial gravity because the acceleration of an object moving in constant
circular motion causing from net external force is called centripetal
acceleration. It defines to the center or seeking the center.
Given the following:
Cylindrical space station
diameter = 275 meters; 137.5
meters for the radius
Standard gravity =
9.80665 m/s²
Using the formula:
w² x r =g
w² = g / r
w² = 9.80665 m/s²
/ 137.5 m
w² = 9.80665 m/s²
/ 137.5 m
w² = 0.0713 s²
Then take the roots
w = 0.267 this is radians per
second / 2 x (3.1416 which is the pi)
w = 0.0424 rps convert to rpm
w = 0.0424 r/s (1minute / 60
seconds)
w = 7.08 x 10⁻⁴ revolutions per minute
Answer:
1.034m/s
Explanation:
We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

Substituting,


Part B)
For the Part B we need to apply conserving momentum equation, this formula is given by,

Where here
is the velocity after the collision.


