Answer:
Since the spring mass system will execute simple harmonic motion the position as a function of time can be written as
'A' is the amplitude = 6 inches (given)
is the natural frequency of the system
At equilibrium we have
Applying values we get
thus natural frequency equals
Thus the equation of motion becomes
At time t=0 since mass is at it's maximum position thus we have
Thus the position of mass at the given times is as follows
1) at
2) at
3) at
4) at
5) at
Answer:
E) are almost circular, with low eccentricities.
Explanation:
Kepler's laws establish that:
All the planets revolve around the Sun in an elliptic orbit, with the Sun in one of the focus (Kepler's first law).
A planet describes equal areas in equal times (Kepler's second law).
The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).
Where T is the period of revolution and a is the semi-major axis.
Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.
However, those orbits have low eccentricities (remember that an eccentricity = 0 corresponds to a circle)
In some moments of their orbit, planets will be closer to the Sun (known as perihelion). According with Kepler's second law to complete the same area in the same time, they have to speed up at their perihelion and slow down at their aphelion (point farther from the Sun in their orbit).
Therefore, option A and B can not be true.
In the celestial sphere, the path that the Sun moves in a period of a year is called ecliptic, and planets pass very closely to that path.
Answer:
The total charge on the rod is 47.8 nC.
Explanation:
Given that,
Charge = 5.0 nC
Length of glass rod= 10 cm
Force = 840 μN
Distance = 4.0 cm
The electric field intensity due to a uniformly charged rod of length L at a distance x on its perpendicular bisector
We need to calculate the electric field
Using formula of electric field intensity
Where, Q = charge on the rod
The force is on the charged bead of charge q placed in the electric field of field strength E
Using formula of force
Put the value into the formula
We need to calculate the total charge on the rod
Put the value into the formula
Hence, The total charge on the rod is 47.8 nC.
Missing details: figure of the problem is attached.
We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,
where:
is the pressure difference between the two ends
is viscosity of the fluid
L is the length of the pipe
is the volumetric flow rate, with
being the section of the tube and
the velocity of the fluid
r is the radius of the pipe.
We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:
is the dynamic water viscosity at
L=4.0 cm=0.04 m
and r=1 mm=0.001 m
Using these data in the formula, we get:
However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.
Answer:
Maximum height the atmosphere pressure can support the
water=10.336 m
Explanation:
We know that ,
Case 1 - Mercury in the tube
Case 2 - Water in the tube
Since atmospheric pressure is same
.
or,
∴
Hence height of the water column =10.336 m