First of all, what a weight is? weight is m × g.. and when a person is in space then the force of gravity isn't acting on him or her, so yeah it will not have weight, but remember gravity is a very very low, now if a body has more mass, more gravity will act on it right? just imagine the mass of a person and that of earth.. now yes, earth will do exert a force on the astronaut but is negiligable.
The same cycle of phases repeats over and over and over and over and over again, so I could start the list with whatever phase I want, and build the list until I get to the same phase I started with.
But the cultures that track their months by the phases of the Moon (traditional Muslim, traditional Jewish, traditional Chinese) all start the new month with the New Moon, so I guess I'll start my list there too.
-- New Moon
-- Waxing Crescent
-- First Quarter
-- Waxing Gibbous
-- Full Moon
-- Waning Gibbous
-- Third Quarter
-- Waning Crescent
-- next New Moon
If a capacitor's dielectric constant is vacuum its dielectric constant is k will be equal to 1.
<u>Explanation:</u>
Relative permittivity of a dielectric substance is referred to as its dielectric constant. Relative permittivity/dielectric constant k is a dimensionless quantity that is the ratio of absolute permittivity and vacuum permittivity.
It is given by the expression
k=k=ε /ε0
where ε denotes absolute permittivity and ε0 denotes permittivity of vacuuum.
Absolute permittivity ε of vacuum= ε0
therefore k= ε0/ ε0=1
dielectric constant of vacuum is 1 .
Answer:
Explanation:
First ship starts at Noon with speed 20 Knots towards West
now we know that 2nd ship starts at 6 PM with speed 15 Knots towards North West
so after time "t" of 2nd ship motion the two ships positions are given as
now we can find the distance between two ships as
now we have
now we will differentiate it with respect to time
here we know that
so we have
now we have
Answer:
I = 0.002593 A = 2.593 mA
Explanation:
Current density = J = (3.00 × 10⁸)r² = Br²
B = (3.00 × 10⁸) (for ease of calculations)
The current through outer section is given by
I = ∫ J dA
The elemental Area for the wire,
dA = 2πr dr
I = ∫ Br² (2πr dr)
I = ∫ 2Bπ r³ dr
I = 2Bπ ∫ r³ dr
I = 2Bπ [r⁴/4] (evaluating this integral from r = 0.900R to r = R]
I = (Bπ/2) [R⁴ - (0.9R)⁴]
I = (Bπ/2) [R⁴ - 0.6561R⁴]
I = (Bπ/2) (0.3439R⁴)
I = (Bπ) (0.17195R⁴)
Recall B = (3.00 × 10⁸)
R = 2.00 mm = 0.002 m
I = (3.00 × 10⁸ × π) [0.17195 × (0.002⁴)]
I = 0.0025929449 A = 0.002593 A = 2.593 mA
Hope this Helps!!!
