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malfutka [58]
3 years ago
10

A solid conducting sphere of radius 2.00 cm has a charge of 8.84 μC. A conducting spherical shell of inner radius 4.00 cm and ou

ter radius 5.00 cm is concentric with the solid sphere and has a charge of −2.02 μC. Find the electric field at the following radii from the center of this charge configuration.
Physics
1 answer:
AlladinOne [14]3 years ago
6 0

Answer: The electric field is given in three regions well defined; 0<r<2; 2<r<4; 4<r<5 and r>5

Explanation:  In order to solve this problem  we have to use the gaussian law in the mentioned regions.

Region 1; 0<r<2

∫E.ds=Qinside the gaussian surface/ε0

inside of the solid conducting sphere the elevctric field is zero because the charge is located at the surface on this sphere.

Region 2; 2<r<4;

E.4*π*r^2=8,84/ε0

E=8,84/(4*π*ε0*r^2)

Region 3; 4<r<5

E=0 because is inside the conductor.

Finally

Region 4; r>5

E.4*π*r^2=(8,84-2.02)/ε0

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A ball is thrown horizontally from the top of a 60 m building and lands 100 m from the base of the building. How long is the bal
zhannawk [14.2K]

Answer:

The ball is in the air for 3.5 seconds

The initial horizontal component of velocity is 28.6 m/s

The vertical component of the final velocity is 34.3 m/s downward

The final velocity is 44.7 m/s in the direction 50.2° below the horizontal

Explanation:

A ball is thrown horizontally

That means the vertical component of the initial velocity u_{y}=0

The initial velocity is the horizontal component u_{x}

The ball is thrown from the top of a 60 m

That means the vertical displacement component y = 60 m

→ y = u_{y} t + \frac{1}{2} gt²

where g is the acceleration of gravity and t is the time

y = -60 m , g = -9.8 m/s² , u_{y}=0

Substitute these values in the rule

→ -60 = 0 + \frac{1}{2} (-9.8)t²

→ -60 = -4.9t²

Divide both sides by -4.9

→ 12.2449 = t²

Take √ for both sides

∴ t = 3.5 seconds

* <em>The ball is in the air for 3.5 seconds </em>

The initial velocity is the horizontal component u_{x}

The ball lands 100 meter from the base of the building

That means the horizontal displacement x = 100 m

→ x = u_{x} t

→ t = 3.5 s , x = 100 m

Substitute these values in the rule

→ 100 = u_{x} (3.5)

Divide both sides by 3.5

→ u_{x} = 28.57 m/s

<em>The initial horizontal component of velocity is 28.6 m/s</em>

The vertical component of the final velocity is v_{y}

→ v_{y} = u_{y} + gt

→ u_{y} = 0 , g = -9.8 m/s² , t = 3.5 s

Substitute these values in the rule

→ v_{y} = 0 + (-9.8)(3.5)

→ v_{y} = -34.3 m/s

<em>The vertical component of the final velocity is 34.3 m/s downward</em>

The final velocity v is the resultant vector of  v_{x} and v_{y}

→ Its magnetude is v=\sqrt{(v_{x})^{2}+(v_{y})^{2}}

→ Its direction tan^{-1}\frac{v_{y}}{v_{x}}

→ v_{y} = 28.6 , v_{y} = -34.3

Substitute this values in the rules above

→ v=\sqrt{(28.6)^{2}+(-34.3)^{2}}=44.66

→ Its direction tan^{-1}\frac{-34.3}{28.6}=-50.18

The negative sign means the direction is below the horizontal

<em>The final velocity is 44.7 m/s in the direction 50.2° below the horizontal</em>

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