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3 years ago

9

Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the magnitude

of the electric field strength between them, if the potential 7.35 cm from the zero volt plate (and 2.65 cm from the other) is 533 V

1 answer:

aivan3 [116]3 years ago

7 0

Answer:

$E=6.8Kv/m$

Explanation:

From the question we are told that

Distance b/w plate $d=10cm=>0.1m$

P_1 Potential at 7.35 $V=533v$

Generally the equation for electric field at a distance is mathematically given as

$E=\frac{v}{d}$

$E=\frac{533}{7.85*10^-^2}$

$E=6789.808917$

$E=6.8*10^3$

$E=6.8Kv/m$

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2 years ago

Can some answer 7 a and b please

pav-90 [236]

Answer:

a.After $15$ second Mr Comer's speed $=$ $1.7 m/s$

b.Distance travelled by Mr.Comer in $15$ seconds $=$ $18.75\ m$

Explanation:

a. Lets recall our first equation of motion $V_{f} =V_{i} + at$

Now we know that $V_{i} = 0.8m/s$ , $a = 0.06\ m/s^2$ and

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$V_{f} =V_{i} + at$

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b.

Lets find the distance and recall our third equation of motion.

$V_{f} ^2-V{i}^2 = 2as$

So $s =$ distance covered.

Dividing both sides with 2a we have.

$\frac{V_{f} ^2-V{i}^2}{2a} = 2as$

Plugging the values.

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3 years ago

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Answer:

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$v_{f}$ = Final speed of rock = 392 ms⁻¹

$m$ = mass of the rock

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Using conservation of energy

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