Answer:
the force required to accelerate a 1,100kg car is 550N
A solar eclipse will be visible over a wide area of the north polar region
on Friday, March 20.
England is not in the path of totality, but it's close enough so that a large
part of the sun will be covered, and it will be a spectacular sight.
For Londoners, the eclipse begins Friday morning at 8:25 AM,when the
moon just begins to eat away at the sun's edge. It advances slowly, as more
and more of the sun disappears, and reaches maximum at 9:31 AM. Then
the obscured part of the sun begins to shrink, and the complete disk is
restored by the end of the eclipse at 10:41AM, after a period of 2 hours
16 minutes during which part of the sun appears to be missing.
The catch in observing the eclipse is:
<em><u>YOU MUST NOT LOOK AT THE SUN</u></em>.
Staring at the sun for a period of time can cause permanent damage to
your vision, even though <em><u>you don't feel it while it's happening</u></em>.
This is not a useful place to try and give you complete instructions or
suggestions for observing the sun over a period of hours. Please look
in your local newspaper, or search online for phrases like "safe eclipse
viewing".
Answer:
1.24 C
Explanation:
We know that the magnitude of the induced emf, ε = -ΔΦ/Δt where Φ = magnetic flux and t = time. Now ΔΦ = Δ(AB) = AΔB where A = area of coil and change in magnetic flux = Now ΔB = 0 - 0.750 T = -0.750 T, since the magnetic field changes from 0.750 T to 0 T.
The are , A of the circular loop is πD²/4 where D = diameter of circular loop = 16.7 cm = 16.7 × 10⁻²m
So, ε = -ΔΦ/Δt = -AΔB/Δt= -πD²/4 × -0.750 T/Δt = 0.750πD²/4Δt.
Also, the induced emf ε = iR where i = current in the coil and R = resistance of wire = ρl/A where ρ = resistivity of copper wire =1.68 × 10⁻⁸ Ωm, l = length of wire = πD and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m.
So, ε = iR = iρl/A = iρπD/πd²/4 = 4iρD/d²
So, 4iρD/d² = 0.750πD²/4Δt.
iΔt = 0.750πD²/4 ÷ 4iρD/d²
iΔt = 0.750πD²d²/16ρ.
So the charge Q = iΔt
= 0.750π(Dd)²/16ρ
= 0.750π(16.7 × 10⁻²m 2.25 × 10⁻³ m)²/16(1.68 × 10⁻⁸ Ωm)
= 123.76 × 10⁻² C
= 1.2376 C
≅ 1.24 C
A green liquid becoming a red liquid
