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3 years ago

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Fariza wears a red hat in her school play. On stage, she is lit by a spotlight shining only green light. When our eyes receive n

o light from an object, it looks black. Explain why Fariza's hat looks black on stage.

1 answer:

FromTheMoon [43]3 years ago

5 0

Answer:It's how the color mixes together . Red and green both makes a dark color

Explanation:

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3 years ago

Difference between artificial magnet and natural magnet??

marissa [1.9K]

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3 years ago

Read 2 more answers

A current is established in a gas discharge tube when a sufficiently high potential difference is applied across the two electro

pishuonlain [190]

Explanation:

It is given that the number of electrons passing through the cross-sectional area in 1 s is $3.4 \times 10^{18}$ . Also, we know that charge on an electron is $-1.60 \times 10^{-19} C$ , then negative charge crossing to the left per second is as follows.

I- = $3.4 \times 10^{18} electrons \times -1.6 x 10^{-19} C/electrons$

I- = 0.544 A

As it is given that the number of protons crossing per second is $1.4 \times 10^{18}$ , as the charge on the proton is $+1.60 \times 10^{-19} C$ , then positive charge crossing to the right per second is calculated as follows.

I+ = $1.4 \times 10^{18} electrons \times 1.6 \times 10^{-19} electrons/C$

I+ = 0.224 A

I = l I+ l + l I- l

So, I = 0.544 + 0.224

= 0.768 A

Thus, we can conclude that the current in given hydrogen discharge tube is 0.768 A.

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3 years ago

If an electric wire is allowed to produce a magnetic field no larger than that of the earth (0.50×10−4t) at a distance of 12 cm

Studentka2010 [4]

The magnetic field generated by a wire carrying a current I is:
$B(r) = \frac{\mu_0 I}{2 \pi r}$
where r is the distance at which the magnetic field is measured, and $\mu_0 = 4 \pi \cdot 10^{-7} NA^{-2}$ is the magnetic permeability in vacuum.

The problem says that the magnetic field at a distance r=12 cm=0.12 m from the wire must be no larger than $B=0.5 \cdot 10^{-4}T$ . Substituting these values, we can find the maximum value of the current I that the wire can carry:
$I= \frac{2 \pi r B}{\mu _0} = \frac{2 \pi (0.12 m)(0.5 \cdot 10^{-4}T)}{ 4 \pi \cdot 10^{-7} NA^{-2}}= 30 A$

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3 years ago