Answer:
Mass of Jupiter = 4.173×10^15kg
Explanation:
Using Kepler's 3rd law, it states that the orbital period T is related to the distance,r as:
T^2 = GM/4 pi × r^3
Where G = universal gravitational constant
r = radius
M = masd of jupiter
Rearranging the formular to make M the subject of formular
T^2 × 4 pi = G M × r^3
(T^2 × 4 pi) / (G× r^3) = M
(1.24^2 × 4 × 3.142) /(6.672×10^-11)(4.11×10^8)^3
M = 19.32 /6.672×10^-11)(4.11×10^8)^3
M = 19.32 / 4.63 ×10^15
M = 4.173×10^15kg
Answer:
I belive it would be ture
Explanation:
It's been a while since I learned this but I think that is right.
Answer:
Explanation:
Given that,
Assume number of turn is
N= 1
Radius of coil is.
r = 5cm = 0.05m
Then, Area of the surface is given as
A = πr² = π × 0.05²
A = 7.85 × 10^-3 m²
Resistance of
R = 0.20 Ω
The magnetic field is a function of time
B = 0.50exp(-20t) T
Magnitude of induce current at
t = 2s
We need to find the induced emf
This induced voltage, ε can be quantified by:
ε = −NdΦ/dt
Φ = BAcosθ, but θ = 90°, they are perpendicular
So, Φ = BA
ε = −NdΦ/dt = −N d(BA) / dt
A is a constant
ε = −NA dB/dt
Then, B = 0.50exp(-20t)
So, dB/dt = 0.5 × -20 exp(-20t)
dB/dt = -10exp(-20t)
So,
ε = −NA dB/dt
ε = −NA × -10exp(-20t)
ε = 10 × NA exp(-20t)
Now from ohms law, ε = iR
So, I = ε / R
I = 10 × NA exp(-20t) / R
Substituting the values given
I = 10×1× 7.85 ×10^-3×exp(-20×2)/0.2
I = 1.67 × 10^-18 A
Answer:
The Sun has a north and south pole, just as the Earth does, and rotates on its axis. However, unlike Earth which rotates at all latitudes every 24 hours, the Sun rotates every 25 days at the equator and takes progressively longer to rotate at higher latitudes, up to 35 days at the poles. This is known as differential rotation.
Explanation:
Answer:
please find attached file
Explanation:
