Answer:
A). σ = 3.823 x
/N-
B).
C/
C).
J
Explanation:
A). We know magnitude of charge per unit area for a conducting plate is given by

where, E is resultant electric field = 1.2 x
V/m
is permittivity of free space = 8.85 x
/N-
k is dielectric constant = 3.6
∴
= 3.6 x 8.85 x
x 1.2 x 
= 3.823 x
/N-
B).Now we know that the magnitude of charge per unit area on the surface of the dielectric plate is given by


C/
C).
Area of the plate, A = 2.5 
= 2.5 x 

diameter of the plate, d = 1.8 mm
= 1800 m
∴ Total energy stored in the capacitor


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Rw^2 = GmM/r^2
<span> Leads to
</span><span> w^2 r^3 = GM
</span><span> (2pi /T) ^2 r^3 = GM
</span><span> 4pi^2 r^3 = GM T^2
</span><span> r^3 = GM T^2 / 4pi^2
</span><span> Work out r^3 then r.
</span> T = 125 min = 125(60) = 7500 s
<span> R = 6.38E6 m
</span><span> m = 5.97E24 kg
</span><span> G = 6.673E-11
</span> r=<span>
8279791.78</span><span> m
Since r = radius R of Earth + height above urface,h
</span><span> h = r - R = </span><span>
8279791.78 - </span>6.38E6 = <span>
<span>1899791.78 m
h=</span></span><span>
<span>1899.79178 Km</span></span>
Answer: Taxpayers
Explanation:
Taking alcohol before driving or while driving is dangerous and has resulted in lots of accidents and deaths. Alcohol tampers with the normal functioning of the brain, and also impairs ones reasoning.
Alcohol impaired driving cases handled by government officials are paid for by the taxpayers. A tax is the levy that the people in the country pays. Those funds are used in handling different government objectives and this is one of such ways.
<span>an object that appears black absorbs al color. an object that appears white reflects all colors.</span>
Answer:
h = 2 R (1 +μ)
Explanation:
This exercise must be solved in parts, first let us know how fast you must reach the curl to stay in the
let's use the mechanical energy conservation agreement
starting point. Lower, just at the curl
Em₀ = K = ½ m v₁²
final point. Highest point of the curl
= U = m g y
Find the height y = 2R
Em₀ = Em_{f}
½ m v₁² = m g 2R
v₁ = √ 4 gR
Any speed greater than this the body remains in the loop.
In the second part we look for the speed that must have when arriving at the part with friction, we use Newton's second law
X axis
-fr = m a (1)
Y Axis
N - W = 0
N = mg
the friction force has the formula
fr = μ N
fr = μ m g
we substitute 1
- μ mg = m a
a = - μ g
having the acceleration, we can use the kinematic relations
v² = v₀² - 2 a x
v₀² = v² + 2 a x
the length of this zone is x = 2R
let's calculate
v₀ = √ (4 gR + 2 μ g 2R)
v₀ = √4gR( 1 + μ)
this is the speed so you must reach the area with fricticon
finally have the third part we use energy conservation
starting point. Highest on the ramp without rubbing
Em₀ = U = m g h
final point. Just before reaching the area with rubbing
= K = ½ m v₀²
Em₀ = Em_{f}
mgh = ½ m 4gR(1 + μ)
h = ½ 4R (1+ μ)
h = 2 R (1 +μ)