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marshall27 [118]

3 years ago

11

Cold plasma used in TVs is different from the solar plasma because ___________.

1 answer:

Arisa [49]3 years ago

5 0

Answer: B i believe

Explanation:

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Which of the following is true for a parallel circuit?

siniylev [52]

B. The voltage is the same across all resistors in the circuit.

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We sometimes conform to others because they belong to a certain ____

Andrew [12]

A.) reference group

"A reference group includes individuals or groups that influence our opinions, beliefs, attitudes and behaviors. They often serve as our role models and inspiration"(study.com).

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During lightning strikes from a cloud to the ground, currents as high as 2.50×10^4 Amps can occur and last for about 40.0 micros

dangina [55]

Answer:

1 C

Explanation:

The intensity of electric current is defined as

$I=\frac{q}{t}$

where

I is the current

q is the amount of charge transferred

t is the time interval during which the charge is transferred

For the lightning in this problem, we have

$I=2.50\cdot 10^4 A$ is the current

$t=40.0 \mu s = 40.0\cdot 10^{-6} s$ is the time interval

Solving the formula for q, we find the amount of charge transferred:

$q=I t = (2.50\cdot 10^4 A)(40.0\cdot 10^{-6}s)=1 C$

6 0

3 years ago

Hàng ngày ta thấy Mặt trời , Mặt trăng quay quanh Trái Đất, khi đó ta đã lấy vật mốc là

marshall27 [118]

I’m sorry I don’t understand this language

5 0

3 years ago

Four traveling waves are described by the following equations, where all quantities are measured in SI units and y represents th

agasfer [191]

Answer:

$T_1=T_3=\dfrac{2\pi}{21}$

$T_2=T_4=\dfrac{2\pi}{42}$

Explanation:

Wave 1, $y_1=0.12\ cos(3x-21t)$

Wave 2, $y_2=0.15\ sin(6x+42t)$

Wave 3, $y_3=0.13\ cos(6x+21t)$

Wave 4, $y_4=-0.27\ sin(3x-42t)$

The general equation of travelling wave is given by :

$y=A\ cos(kx\pm \omega t)$

The value of $\omega$ will remain the same if we take phase difference into account.

For first wave,

$\omega_1=21$

$\dfrac{2\pi }{T_1}=21$

$T_1=\dfrac{2\pi}{21}$

For second wave,

$\omega_2=42$

$\dfrac{2\pi }{T_2}=42$

$T_2=\dfrac{2\pi}{42}$

For the third wave,

$\omega_3=21$

$\dfrac{2\pi }{T_3}=21$

$T_3=\dfrac{2\pi}{21}$

For the fourth wave,

$\omega_4=42$

$\dfrac{2\pi }{T_4}=42$

$T_4=\dfrac{2\pi}{42}$

It is clear from above calculations that waves 1 and 3 have same time period. Also, wave 2 and 4 have same time period. Hence, this is the required solution.

3 0

3 years ago