The center of the Hubble space telescope is 6940 km from Earth’s center. If the gravitational force between Earth and the telesc
2 answers:
The gravitational attraction between the Earth and the Hubble telescope is given by:
where
G is the gravitational constant
M is the Earth's mass
m is the Hubble mass
r is the distance of the Hubble telescope from the earth's center
By rearranging the equation and substituting the numbers given by the problem, we find the mass of the telescope:
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Answer:
F = 63N
Explanation:
M= 1.5kg , t= 2s, r = (2t + 10)m and
Θ = (1.5t² - 6t).
magnitude of the resultant force acting on 1.5kg = ?
Force acting on the mass =
∑Fr =MAr
Fr = m(∇r² - rθ²) ..........equation (i)
∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)
The horizontal path is defined as
r = (2t + 10)
dr/dt = 2, d²r/dt² = 0
Angle Θ is defined by
θ = (1.5t² - 6t)
dθ/dt = 3t, d²θ/dt² = 3
at t = 2
r = (2t + 10) = (2*(2) +10) = 14
but dr/dt = 2m/s and d²r/dt² = 0m/s
θ = (1.5(2)² - 6(2) ) = -6rads
dθ/dt =3(2) - 6 = 0rads
d²θ/dt = 3rad/s²
substituting equation i into equation ii,
Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)
∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]
∑F = 1.5(14*3+0) = 63N
F = √(Fr² +FΘ²) = √(0² + 63²) = 63N
Answer:
Explanation:
It is given that,
A planar electromagnetic wave is propagating in the +x direction.The electric field at a certain point is, E = 0.082 V/m
We need to find the magnetic vector of the wave at the point P at that instant.
The relation between electric field and magnetic field is given by :
c is speed of light
B is magnetic field
So, the magnetic vector at point P at that instant is .