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marta [7]
3 years ago
6

A flat sheet of paper of area 0.250 m2 is oriented so that the normal to the sheet is at an angle of 60 to a uniform electric fi

eld of magnitude 14 N>C. (a) Find the magnitude of the electric flux through the sheet. (b) Does the answer to part (a) depend on the shape of the sheet
Physics
1 answer:
Harman [31]3 years ago
4 0

Answer:

a. 1.75 Nm²/C

b. Yes.

Explanation:

a. Electric Flux is given as:

Φ = E*A*cosθ

Where E = electric flux

A = Surface area

Φ = 14 * 0.25 * cos60

Φ = 1.75 Nm²/C

b. Yes, the shape of the sheet will affect the Flux through it. This is because flux is dependent on area of the surface and the area is dependent on the shape of the surface.

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M1U1 + M2V2 = (M1+M2)V, where M1 is the mass of the moving car, M2 is the mass of the stationary car, U1 is the initial velocity, and V is the common velocity after collision.
therefore; 
(1060× 16) + (1830 ×0) = (1060 +1830) V
 16960 = 2890 V
      V = 5.869 m/s
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If the chain reaction is not controlled, what can happen?
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2 years ago
A 6.00 kg ball is dropped from a height of 12.0 m above one end of a uniform bar that pivots at its center. The bar has mass 5.0
Andre45 [30]

Answer:

The height of the other ball go after the collision is 2.304 m.

Explanation:

Given that,

Mass of ball = 6.00 kg

Height = 12.0 m

Mass of bar =5.00 kg

Length = 4.00 m

Suppose we need to calculate how high will the other ball go after the collision

We need to calculate the velocity of ball

Using formula of velocity

v=\sqrt{2gh}

v=\sqrt{2\times9.8\times12.0}

v=15.33\ m/s

We need to calculate the angular momentum

Using formula of angular momentum

l_{before}=mvr

Put the value into the formula

l_{before}=6.00\times15.33\times2.0

l_{before}=183.96\ kgm^2/s

We need to calculate the angular momentum

Using formula of angular momentum

l_{after}=I_{t}\omega

l_{after}=(\dfrac{ml^2}{12}+m_{1}r^2+m_{2}r^2)\omega

Put the value into the formula

l_{after}=(\dfrac{5\times4.00^2}{12}+6.00\times2.0^2+6.00\times2.0^2)\omega

183.96=54.66\omega

\omega=\dfrac{183.96}{54.66}

\omega=3.36\ rad/sec

After collision the ball leaves with velocity

We need to calculate the velocity after collision

Using formula of the velocity

v= r\omega

v=2.0\times3.36

v=6.72\ m/s

We need to calculate the height

Using formula of height

h=\dfrac{v^2}{2g}

Put the value into the formula

h=\dfrac{(6.72)^2}{2\times9.8}

h=2.304\ m

Hence, The height of the other ball go after the collision is 2.304 m.

5 0
3 years ago
Why does helium exhibit more visible emission lines than hydrogen?
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