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il63 [147K]
3 years ago
8

How does the electric field intensity vary with the increase of distance of the point from the centre of a charged conducting sp

here? Explain with graph.​
Physics
1 answer:
Nataly_w [17]3 years ago
4 0

Answer:

(i)  Electric field outside the shell:

For point r>R; draw a spherical gaussian surface of radius r.

Using gauss law, ∮E.ds=q0qend

Since E is perpendicular to gaussian surface, angle betwee E is 0.

Also E being constant, can be taken out of integral.

So, E(4πr2)=q0q

So, E=4πε01r2q

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In which of these examples is the greatest movement occurring?
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not clear pic...but it's definitely not A)

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Which statement is a scientific theory?
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A.

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A sky diver with a mass of 70kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be Fd=kV^
xeze [42]

Answer:

v_{max}=52.38\frac{m}{s}

v_{100}=33.81

Explanation:

the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:

\sum{F}=0=F_d-W

F_d=W

kv_{max}^2=m*g

v_{max}=\sqrt{\frac{m*g}{k}} =\sqrt{\frac{70*9.8}{0.25}}=52.38\frac{m}{s}

To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:

\sum{F}=ma=W-F_d

ma=W-F_d

ma=mg-kv_{100}^2

a=g-\frac{kv_{100}^2}{m} (1)

consider the next equation of motion:

a = \frac{(v_{x}-v_0)^2}{2x}

If assuming initial velocity=0:

a = \frac{v_{100}^2}{2x} (2)

joining (1) and (2):

\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}

\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g

v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g

v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}

v_{100}=\sqrt{\frac{g}{(\frac{1}{2x}+\frac{k}{m})}} (3)

v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}

v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}

v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}

v_{100}=\sqrt{1,143.3}

v_{100}=33.81

To plot velocity as a function of distance, just plot equation (3).

To plot velocity as a function of time, you have to consider the next equation of motion:

v = v_0 +at

as stated before, the initial velocity is 0:

v =at (4)

joining (1) and (4) and reducing you will get:

\frac{kt}{m}v^2+v-gt=0

solving for v:

v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }

Plots:

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3 years ago
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