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3 years ago

The visible light spectrum ranges between

Physics

2 answers:

natulia [17]3 years ago

8 0

Answer: 380 nm (violet-blue) and 780 nm (red)

Explanation:

The portion visible by the human eye of the electromagnetic spectrum is between 380 nm (violet-blue) and 780 nm (red) approximately.

This part of the spectrum is located between ultraviolet light and infrared light.

It should be noted that the fact only part of the whole electromagnetic spectrum is visible to humans is because the receptors in our eyes are only sensitive to these wavelengths.

aleksandr82 [10.1K]3 years ago

4 0

Red goes in the first blank and violet goes in the second blank if you're using plato

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Now let’s apply Coulomb’s law and the superposition principle to calculate the force on a point charge due to the presence of ot

Mnenie [13.5K]

Answer:

F = - 1.68 10⁻⁴ N

, it is directed to the left of the x-axis

Explanation:

Coulomb's law is

F = k q₁ q₂ / r²

Where K is the Coulomb constant that value 8.99 10⁹ N m²/ C², q are the electric charges and r is the distance between them. Let's apply to our problem for each pair of charges

Let's reduce the magnitudes to the SI system

q₁ = 3.0 nc (1C / 10 9 nC) = 3.0 10⁻⁹ C

x₁ = 2.0 cm (1m / 100cm) = 2.0 10⁻² m

q₂ = -6.0 nC = -6.0 10⁻⁹ C

x₂ = 4.0 cm = 4.0 10⁻² m

q₃ = 5.0 nC = 5.0 10⁻⁹ C

x3 = 0 m

Charges q1 and q3

r = x₁ -x₃

r = 2.0 10⁻² -0

r = 2.0 10⁻² m

F₁₃ = 8.99 10⁹ 3.0 10⁻⁹ 5.0 10⁻⁹ / (2.0 10⁻²)²

F₁₃ = 33.7 10⁻⁵ N

As the charges are of the same sign, the force is repulsive, therefore it is directed to the left of the x-axis

Charges q2 and q3

r = r₂ –r₃

r = 4.0 10⁻² - 0 = 4.0 10⁻² m

F₂₃ = 8.99 10⁹ 6.0 10⁻⁹ 5.0 10⁻⁹ / (4.0 10⁻²)²

F₂₃ = 16.86 10⁻⁵ N

As the charges are of different sign, the force is attractive, therefore it is directed to the right of the x-axis

The force is a vector magnitude, so each component must be added independently, in this case all the forces are on the x-axis, let's take the right direction as positive

F = F₂₃ - F₁₃

F = 16.86 10⁻⁵ - 33.7 10⁻⁵

F = - 16.84 10⁻⁵ N

F = - 1.68 10⁻⁴ N

The negative sign means that it is directed to the left of the x-axis

5 0

3 years ago

The illuminance due to a 60.0-w lightbulb at 3.0m is 9.35 lx. What is the total luminous flux of the bulb

Furkat [3]

The equation for luminous flux is given as P = 4 $\pi$ $r^{2}$ E
where P is the luminous flux, r is the distance and E is the illumination. The unit for P is lumen, E is lux and r is in meters. Substituting the given to the equation:

P = 4 $\pi$ $r^{2}$ E

P= 4 $\pi$ $(3)^{2}$ (9.35) = 1057.46 lumens (lm)

The total luminous flux is equal to 1057.46 lumens (lm).

3 0

3 years ago

A gas was compressed to 30.0 mL at 1.5 atm from 65 mL. What was the original pressure?

astra-53 [7]

A gas as 30.0 Ml because pile of 1.4 add 1 that equals. To 1.5 and remove 10 so thebpresssure is 1399.93

4 0

3 years ago

Just about everyone at one time or another has been burned by hot water or steam. This problem compares the heat input to your s

tatyana61 [14]

Answer:

$Q_T=63313.5\ J$

Explanation:

Given:

temperature of skin, $T_s=34^{\circ}C$

initial temperature of steam vapour, $T_v=100^{\circ}C$

latent heat of steam, $L=2256\ J.g^{-1}$

mass of steam, $m=25\ g$

specific heat of water, $c=4190\ J.kg^{-1}.K^{-1}=4.19\ J.g^{-1}.K^{-1}$

final temperature, $T_f=34^{\circ}C$

Assuming that no heat is lost in the surrounding.

We know:

$Q=m.c.\Delta T$

Now the total heat given by the steam to form water at the given conditions:

$Q_T=Q_{Lv}+Q_w$ ..............................(1)

where:

$Q_{Lv}=$ latent heat given out by vapour to form water of 100°C

$Q_w=$ heat given by water of 100°C to come at 34°C.

putting respective values in eq. (1)

$Q_T=m(L+c.\Delta T)$

$Q_T=25(2256+4.19\times 66)$

$Q_T=63313.5\ J$

is the heat transferred to the skin.

4 0

3 years ago

A metal rod A and a metal sphere B, on insulating stands, touch each other. They are originally neutral. A positively charged ro

ryzh [129]

Answer:

The sphere is positively charged

Explanation:

This is because when the positively charged rod is brought near the metal rod A, the electrons in metal rod A and sphere B are attracted towards it into metal rod A while the positive charges in the are repelled into sphere B. So, when the charged rod is withdrawn, and metal rod A and sphere B are separated, metal rod A is now negatively charged, but sphere B is positively charged.

So, sphere B is positively charged.

3 0

3 years ago