Answer:
[1, 6, -2]
Explanation:
Given the following :
Initial Position of spaceship : [3 2 4] km
Velocity of spaceship : [-1 2 - 3] km/hr
Location of ship after two hours have passed :
Distance moved by spaceship :
Velocity × time
[-1 2 -3] × 2 = [-2 4 -6]
Location of ship after two hours :
Initial position + distance moved
[3 2 4] + [-2 4 -6] = [3 + (-2)], [2 + 4], [4 + (-6)]
= [3-2, 2+4, 4-6] = [1, 6, -2]
the awnser to ur question is D
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
<em>Important thing is that all unitless quantity is dimensionless quantity. .</em><em>A</em><em> dimensionless physical quantity may have an unit</em>
Answer:
it will take 36.12 ms to reduce the capacitor's charge to 10 μC
Explanation:
Qi= C×V
then:
Vi = Q/C = 30μ/20μ = 1.5 volts
and:
Vf = Q/C = 10μ/20μ = 0.5 volts
then:
v = v₀e^(–t/τ)
v₀ is the initial voltage on the cap
v is the voltage after time t
R is resistance in ohms,
C is capacitance in farads
t is time in seconds
RC = τ = time constant
τ = 20µ x 1.5k = 30 ms
v = v₀e^(t/τ)
0.5 = 1.5e^(t/30ms)
e^(t/30ms) = 10/3
t/30ms = 1.20397
t = (30ms)(1.20397) = 36.12 ms
Therefore, it will take 36.12 ms to reduce the capacitor's charge to 10 μC.
