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svp [43]
4 years ago
15

Where is the electric field closest to uniform? inside a circle described by a point charge at its center between parallel plate

s of equal charge and equal magnitude inside a circle described by identical, evenly-spaced point charges around the circumference between parallel plates of equal charge and opposite magnitude
Physics
1 answer:
Soloha48 [4]4 years ago
5 0

1. Inside a circle described by a point charge at its center will not be uniform field as a point charge field will change with distance where as we move away the field strength will go down.

2. between parallel plates of equal charge and equal magnitude will have zero electric field because both plates are of same charge so electric field of both plates will cancel each other.

3. inside a circle described by identical, evenly-spaced point charges around the circumference field will be zero as all charges are distributed uniformly and they cancel the field effect of each other.

4. between parallel plates of equal charge and opposite magnitude field will be nearly uniform as the two charges are opposite in nature and equal in magnitude so the effect will be added.

So answer of the above will be between parallel plates of equal charge and opposite magnitude

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A thin rod of length 0.75 m and mass 0.42 kg is suspended freely from one end. It is pulled to one side and then allowed to swin
Goshia [24]

Answer:

(A) 0.63 J  

(B) 0.15 m

Explanation:

length (L) = 0.75 m

mass (m) =0.42 kg

angular speed (ω) = 4 rad/s

To solve the questions (a) and (b) we first need to calculate the rotational inertia of the rod (I)

I = Ic + mh^{2}  

Ic is the rotational inertia of the rod about an axis passing trough its centre of mass and parallel to the rotational axis

h is the horizontal distance between the center of mass and the rotational axis of the rod

I = (\frac{1}{12})(mL^{2} ) + m([tex]\frac{L}{2})^{2}[/tex]

I = (\frac{1}{12})(0.42 x 0.75^{2} ) + ( 0.42 x ([tex]\frac{0.75}{2})^{2}[/tex])

I = 0.07875 kg.m^{2}

(A) rods kinetic energy = 0.5Iω^{2}

  = 0.5 x 0.07875 x 4^{2} = 0.63 J   0.15 m

(B) from the conservation of energy

   initial kinetic energy + initial potential energy = final kinetic energy + final potential energy

   Ki + Ui = Kf + Uf

   at the maximum height velocity = 0 therefore final kinetic energy = 0

   Ki + Ui = Uf

   Ki = Uf - Ui

 Ki =  mg(H-h)

where (H-h) = rise in the center of mass

     0.63 = 0.42 x 9.8 x (H-h)

   (H-h) = 0.15 m

6 0
3 years ago
A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. how long does h
noname [10]
The shot putter should get out of the way before the ball returns to the launch position.

Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.

The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 =  0.45 s

t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.

Answer: 0.45 s
7 0
3 years ago
it is about 384,750 kilometers from earth to the moon. it took the apollo astronauts about 2 days and 19.5 hours to fly to the m
Jet001 [13]
We know, speed = Distance / Time
d = 384,750 Km
t = 2 days, 19.5 hours = 48+19.5 = 67.5 hour

Substitute their values, 
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In short, Your Answer would be 5700 Km/h

Hope this helps!
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Where:

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Replacing:

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7 0
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pav-90 [236]

Answer:

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Explanation:

6 0
3 years ago
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