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3 years ago

What is the approximate range of wavelengths for visible light?

1 answer:

7 0

We can answer the question by looking at the spectrum of the visible light, for example in this figure:
https://en.wikipedia.org/wiki/Light#/media/File:EM_spectrum.svg

We see from the figure that the lowest part of the spectrum (violet) corresponds approximately to a wavelength of 380 nm, while the highest part of the spectrum (red light) corresponds approximately to a wavelength of 750 nm.

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Which reading strategy helps you to read and understand a passage quickly by making use of your previous knowledge?

Molodets [167]

Taking notes,because you can read back on what you learned

4 0

3 years ago

The basic building block of life is

snow_tiger [21]

The basic building block of life is molecules, more specifically macromolecules. There are four different macromolecules which could all be described as the building blocks of life, namely carbohydrates, proteins, nucleic acids and proteins.

8 0

3 years ago

7) Three resistors having resistances of 4.0 Ω, 6.0 Ω, and 10.0 Ω are connected in parallel. If the combination is connected in

viva [34]

Answer:

A, 0.59A

Explanation:

The total resistance in the circuit is the resistances in parallel plus that in series.

Total resistance for those in parallel is;

1/(1/4 +1/6 +1/10) = 1/ (15+10+6 /60)

1/(31/60)= 60/31 ohms

Hence total resistance of the circuit is;

60/31 + 2 = (60+62)/31 = 122/31=3.94 ohms

To calculate the current flowing through the 10ohm resistance we need to know the voltage drop by subtracting the voltage drop in the 2ohm resistance from the total voltage drop.

Voltage drop on the 2 ohm resistance is;

Current on the 2 ohm resistor × 2 ohms

V = I ×R ; I - current

R - resistance

Current drop on the 2ohm resistance is;

Total voltage in the circuit/ total resistance in the circuit

12/3.94= 3.05A

Voltage drop on the 2 ohm resistance;

3.05 × 2 = 6.10volts

Hence voltage drop on the parallel resistance would be ;

12-6.10= 5.90V

Now voltage drop in a parallel circuit is the same hence 5.90v is dropped in each of the parallel resistance.

That said, the current drop on the 10 ohm resistor would be;

5.90/10 = 0.59A

Remember V= I× R so that I = V/R

6 0

3 years ago

A battery does work on electric charges to bring them to a position of higher electric potential energy so that they can flow th

Ipatiy [6.2K]

(Missing question is: which of the following statements are true?)

a) A battery does work on electric charges to bring them to a position of higher electric potential energy so that they can flow through a circuit to a lower potential energy --> TRUE
That's true: the battery does the work to move the charges to one end of its terminal thus creating a potential difference between the two terminals. Then, when it is connected to the circuit, charges start to flow to the terminal at lower potential (through the circuit)

b) the potential difference between the terminals of a battery, when no current flows to an external circuit, is referred to as the terminal voltage --> FALSE
This is false: when no current flows, it is called e.m.f. (electromotive force)

c) the internal resistance of a battery decreases with decreasing temperature
--> TRUE
In fact, the dependence of a resistance with the temperature is:
 $R=R_0[1+\alpha (T-T_0)]$
with $\alpha$ being generally positive, therefore the value of the resistance is proportional to T, and when T decreases, R decreases as well.

d) a battery is a device that produces electricity by transforming chemical energy into electrical energy --> TRUE
That's true: a battery uses chemical reactions to create a potential difference between the two terminals that can be exploited to make charges flowing through a circuit.

7 0

3 years ago

What is the magnitude of the electric field at a distance of 89 cm from a 27 μC charge, in units of N/C?

GuDViN [60]

Answer:

306500 N/C

Explanation:

The magnitude of an electric field around a single charge is calculated with this equation:

$E(r) = \frac{1}{4 \pi *\epsilon 0} \frac{q}{r^2}$

With ε0 = 8.85*10^-12 C^2/(N*m^2)

Then:

$E(0.89) = \frac{1}{4 \pi *8.85*10^-12} \frac{27*10^-6}{0.89^2}$

E(0.89) = 306500 N/C

3 0

3 years ago