Answer:
mass of the neutron star =3.45185×10^26 Kg
Explanation:
When the neutron star rotates rapidly, a material on its surface to remain in place, the magnitude of the gravitational acceleration on the central material must be equal to magnitude of the centripetal acc. of the rotating star.
That is
M_ns = mass odf the netron star.
G= gravitational constant = 6.67×10^{-11}
R= radius of the star = 18×10^3 m
ω = 10 rev/sec = 20π rads/sec
therefore,
= 3.45185... E26 Kg
= 3.45185×10^26 Kg
Answer:
4 A
Explanation:
The relationship between current, voltage and resistance in a circuit is given by Ohm's law:
where
V is the voltage
R is the resistance
I is the current
The equation can also be rewritten as
from which we see that the current is inversely proportional to the resistance, R.
In this problem, the initial current is I = 8 A. Then the resistance is doubled:
R ' = 2R
So the new current is
so the current is halved.
Answer:
a) see attached, a = g sin θ
b)
c) v = √(2gL (1-cos θ))
Explanation:
In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by
Wₓ = m a
W sin θ = m a
a = g sin θ
b) The diagram is the same, the only thing that changes is the angle that is less
θ' = 9/2 θ
c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.
The easiest way to find linear speed is to use conservation of energy
Highest point
Em₀ = mg h = mg L (1-cos tea)
Lowest point
Emf = K = ½ m v²
Em₀ = Emf
g L (1-cos θ) = v² / 2
v = √(2gL (1-cos θ))
