As you make your way to work in Boston just after dawn, you observe that the bay's water level is surprisingly low. Considering this significantly low tide and the dark, clear sky from last night, the moon is currently in its waxing crescent phase. Among the eight lunar phases, the waxing crescent occurs after the new moon phase.
The lunar phases are a result of the moon's orbit positioned between the Earth and the Sun. The moon's surface reflects sunlight, which reaches Earth at varying angles, depending on the moon's orbit position. This creates the perception of the moon appearing to wax and wane.
D) Both perform no work
Explanation:
The work accomplished by a force is defined as:
where
F represents the applied force
d denotes the displacementis the angle highlighted between the applied force and the displacement vectorFrom this formula, it is clear that work is only done when displacement occurs, meaning the object has to move.
In this instance, as the wall is unmoving, the displacement is zero: d = 0, thus no work is performed.
Answer:
Explanation:
The measurement of pressure is indicated as where p denotes the pressure,
signifies density, and h represents height
Given values include pressure , gravity's acceleration
, and height =1.163 m
Answer:
W = -510.98 J
Explanation:
Force = 43 N, 61° SW
Displacement = 12 m, 22° NE
The work done is calculated using:
W = F*d*cos(A)
where A is the angle between the applied force and displacement.
The angle A between the force and displacement is determined as A = 61 + 90 + 22 = 172°
Hence, W = 43 * 12 * cos(172)
This results in W = -510.98 J
The negative result indicates that the work is done contrary to the direction of the force applied.
Answer:
Explanation:
Data provided
initial velocity v₀=20 cm/s at time t=3s
final velocity vf=0 at time t=8 s
Required
Average Acceleration for the interval from 3s to 8s
Solution
Acceleration can be defined as the first derivative of velocity concerning time
Details that are not provided: the problem figure is included.
We can address the exercise by applying Poiseuille's law. This law indicates that for a fluid flowing in a laminar manner within a confined pipe,
where:
represents the pressure difference across the two ends
denotes the viscosity of the fluid
L signifies the length of the pipe
indicates the volumetric flow rate, where
is the cross-sectional area of the tube and
refers to the fluid's velocity
r stands for the pipe's radius.
This law can be utilized for the needle, allowing us to compute the pressure difference between point P and the needle's end. In this scenario, we have:
is the dynamic viscosity of water at
L=4.0 cm=0.04 m
and r=1 mm=0.001 m
Substituting these values into the formula yields:
This pressure difference specifies the value between point P and the needle's termination. As the end of the needle is under atmospheric pressure, the gauge pressure at point P, relative to atmospheric pressure, is exactly 3200 Pa.
