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slava [35]
3 years ago
7

A long line carrying a uniform linear charge density+50.0μC/m runs parallel to and 10.0cm from the surface of a large, flat plas

tic sheet that has a uniform surface charge density of −100μC/m2on one side.Part AFind the location of all points where an α particle would feel no force due to this arrangement of charged objects.L=_____________ m from the line.Part BChoose an appropriate location of these points at a distance calculated in part A.o above the lineo between the line and the sheet.
Physics
1 answer:
IRISSAK [1]3 years ago
6 0

Answer:

The location of the points is at the distance  above the line charge Option A.

Explanation:

Part A

The electric field due to a line of charge is  expressed as :

Eₐ = λ /2πε₀L  …… (1)

while  the electric field due to a non-conducting elastic sheet is expressed as:

Еₙ =   λ /2ε₀  …… (2)

In order for the  particle to experience no force, the electric field at that point is zero.

Equating  (1) and (2) together, we have:.

λ /2πε₀L = σ /2ε₀  

Rearranging the above expression, we have:

L =  λ /πσ

Substituting 50.0 μC/m  for λ, 3.14 for  π  , and 100 μC/ m² for σ, we have:

L=  50.0 μC/m / (3.14)* 100 μC/ m²

L = 0.159m

From the L value obtained above, the location of all points where an  particle would experience no force due to the arrangement of charged objects is L 0.159m  from the line.

Part B  

The electric field is zero at all the points above the line with the distance 0.159m  from the line.

Thus, the location of the points is at the distance  above the line charge Option A.

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bagirrra123 [75]

Answer:

(a) T = 2987.6 k

(b) T = 19986.2 k

Explanation:

The temperature of a star in terms of peak wavelength can be given by Wein's Displacement Law, which is as follows:

T = \frac{0.2898\ x\ 10^{-2}\ m.k}{\lambda_{max}}

where,

T = Radiated surface temperature

\lambda_{max} = peak wavelength

(a)

here,

\lambda_{max} = 970 nm = 9.7 x 10⁻⁷ m

Therefore,

T = \frac{0.2898\ x\ 10^{-2}\ m.k}{9.7\ x\ 10^{-7}\ m}

<u>T = 2987.6 k</u>

(b)

here,

\lambda_{max} = 145 nm = 1.45 x 10⁻⁷ m

Therefore,

T = \frac{0.2898\ x\ 10^{-2}\ m.k}{1.45\ x\ 10^{-7}\ m}

<u>T = 19986.2 k</u>

6 0
3 years ago
We intend to observe two distant equal brightness stars whose angular separation is 50.0 × 10-7 rad. Assuming a mean wavelength
san4es73 [151]

Answer:

13.4cm

Explanation:

According to Rayleigh’s criterion the angular resolution to distinguish two objects is given by:

\theta=1.22\frac{\lambda}{b}

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By doing b the subject of the formula and replacing the values of the angle and wavelength you obtain:

b=1.22\frac{\lambda}{\theta}=1.22\frac{550*10^{-9}m}{50.0*10^{-7}rad}=0.134m=13.4cm

hence, the smallest diameter objective lens is 13.4cm

8 0
3 years ago
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Answer:

Explanation:

a) Power consumption is 4100 J/min / 60 s/min = 68.3 W(atts)

work done raised the potential energy

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3 years ago
When might it be harder to stop a vehicle moving at 30 km/h than one moving at 60 km/h?
rodikova [14]

Answer:

when the momentum of the vehicle moving at 30 km/h is higher than the one from the vehicle moving at 60 km/h

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It's much harder to stop a freight truck moving at 30 km/h than a hot wheels car moving at 60 km/h.

4 0
2 years ago
A 20-Kg child is on a swing attached to 3.0 m-long chains. The child swings back and forth, swinging out to a 60-degree angle. (
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Answer:

 v = 29.4 m / s

Explanation:

For this exercise we can use the conservation of mechanical energy

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energy is conserved

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         v = 2g L (1- cos 60)

 

let's calculate

          v² = 2 9.8 3.0 (1- cos 60)

          v = 29.4 m / s

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3 years ago
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