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Veronika [31]
3 years ago
5

A flat, wide cloud floats horizontally a few kilometers above the surface of Earth. Its lower surface carries a uniform surface

charge density of -2.6×10-4 C/m2 , while there is no excess charge elsewhere in the cloud. What is the magnitude of the electric field, in newtons per coulomb, a few meters below the cloud, under the cloud’s approximate center, due to the charge distribution?
Physics
1 answer:
valentinak56 [21]3 years ago
4 0

Answer:

\frac{kQ}{r^2} r^

Explanation:

Electric field strength= Force/unit charge

E= (kQq/r²)/q ₓ r

where r is the unit vector in the direction of unit charge

E= \frac{kQ}{r^2} r^

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A trumpet player is tuning his instrument by playing an A note simultaneously with the first-chair trumpeter, who has perfect pi
KIM [24]

To develop this problem, it is necessary to apply the concepts related to Beat

The Beat is an acoustic phenomenon that is generated by two sine waves interfering with slightly different frequencies. The beat frequency is equal to the difference in the frequencies of the two original waves:

f_{bat} = |(f_1 \pm f_B)|

Our values are given as

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f_B = 2.1Hz

For the particular case we have two possible frequencies:

1) f_{bat} = |(f_1 - f_B)|

f_{bat} = |(440 - 2.1)|

f_{bat} = 437.9Hz

2) f_{bat} = |(f_B + f_1)|

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4 0
3 years ago
Choose all correct sentences Group of answer choices The power is maximum when the value of the impedance is greater than the va
Illusion [34]

Answer:

True  b and c

Explanation:

In an RLC circuit the impedance is

         Z = \sqrt{[R^{2} + ( (wL)^{2} + (\frac{1}{wC})^{2} ]     }

examine the different phrases..

a) False. The maximum impedance is the value of the resistance  

b) True. Resonance occurs when  

              (wL)² + (1 / wC)² = 0

               w² = 1 / LC

c) True. In resonance the impedance is the resistive part and the power is maximum  

d) False. In resonance the inductive and capacitive part cancel each other out  

e) False. The impedance is always greater outside of resonance, but at the resonance point they are equal

6 0
3 years ago
A 500kg car skids to a stop at a traffic light, leaving behind a 18.25m skid mark as it comes to a rest. Assuming that the car i
Nastasia [14]

Answer:

Coefficient of friction will be 0.587

Explanation:

We have given mass of the car m = 500 kg

Distance s = 18.25 m

Initial velocity of the car u = 14.5 m/sec

As the car finally stops so final velocity v = 0 m/sec

From second equation of motion

v^2=u^2+2as

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We know that acceleration is given by

a=\mu g

5.76=\mu\times  9.81

\mu =0.587

So coefficient of friction will be 0.587

6 0
3 years ago
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