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Natali [406]
4 years ago
12

A uniform electric field exists in a region between two oppositely charged plates. An electron is released from rest at the surf

ace of the negatively charged plate and strikes the surface of the opposite plate, 1.3 cm away, in a time 7.2 times 10-7 s. What is the magnitude of the electric field?
Physics
1 answer:
Law Incorporation [45]4 years ago
6 0

Answer:

E=0.284 N/C

Explanation:

Given that

Distance ,d= 1.3 cm = 0.013 m

Time ,t

t=7.2\times 10^{-7}\ s

Initial velocity of electron u=0 m/s

We know that

d=ut+\dfrac{1}{2}at^2

0.013=0+\dfrac{1}{2}\times a\times(7.2\times 10^{-7}) ^2

a=5.01\times 10^{10}\ m/s^2

We know that

mass of electron,m

m=9.1\times 10^{-31}\ kg

Charge on electron

q=1.6\times 10^{-19}\ C

F= m a=E q

So

E=\dfrac{ma}{q}\ N/C

E=\dfrac{9.1\times 10^{-31}\times 5.01\times 10^{10}}{1.6\times 10^{-19}}\ N/C

E=0.284 N/C

Electric field will be 0.284 N/C.

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Answer:

0.00016 kg

Explanation:  

Given:

Power = P = 1.2 × 10⁹ Watts

Power =  work done / Time

efficiency = 0.30

Input power = 1.2 × 10⁹ / 0.30 =  4  × 10⁹ W

Energy =  4  × 10⁹ x 60 x 60 = 1.44 x 10¹³ joules

E = m c² , where c is the speed of light and m is the mass.

⇒ mass = m = E / c²  = (1.44 x 10¹³) / (3 × 10⁸ )²

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A vacuum cleaner produces sound with a measured sound level of 75.0 dB. (a) What is the intensity of this sound in W/m2? W/m2 (b
jonny [76]

Answer:

Part a)

I = 3.16 \times 10^{-5} W/m^2

Part b)

P_o = 0.162 Pa

Explanation:

Part a)

Level of sound = 75 dB

now we know that

L = 10 Log\frac{I}{I_0}

here we know that

I_0 = 10^{-12} W/m^2

now we have

75 = 10 Log(\frac{I}{10^{-12}})

I = 3.16 \times 10^{-5} W/m^2

Part b)

Intensity of sound wave is given as

I = \frac{1}{2}\rho A^2\omega^2 c

here we know that

A = \frac{P_o}{Bk}

so we have

I = \frac{1}{2}\rho(\frac{P_o}{Bk})^2\omega^2 c

I = \frac{1}{2}\rho P_o^2 \frac{c^3}{B^2}

now we know

\rho = 1.2 kg/m^3

c = 340 m/s

B = 1.4 \times 10^5 Pa

now we have

3.16 \times 10^{-5} = \frac{1}{2}(1.2)P_o^2\frac{340^3}{(1.4\times 10^5)^2}

P_o = 0.162 Pa

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3 years ago
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