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3 years ago

A television remote control uses infrared light with a wavelength of 940 nm. What is the frequency of the light?

Physics

1 answer:

Ronch [10]3 years ago

5 0

Answer:

Frequency = 3.19 * 10^14 Hz or 1/s

Explanation:

Relationship b/w frequency and wavelength can be expressed as:

C = wavelength * frequency, where c is speed of light in vacuum which is 3.0*10^8 m/s.

Now simply input value (but before that convert wavelength into meters to match the units, you do this by multiply it by 10^-9 so it will be 940*10^-9)

3.0 * 10^8 = Frequency * 940 x 10^-9

Frequency = 3.19 * 10^14 Hz or 1/s

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The higher the pressure on a liquid, the higher its boiling point temperature. true false

Alika [10]

True, p1/t1=p2/t2. Pressure is related to temperature at which it boils so pressure does affect the boiling point.

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3 years ago

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how does the electric force between two charged particles change if the distance between them is increased by a factor of 3? a.

Aleksandr [31]

how does the electric force between two charged particles change if the distance between them is increased by a factor of 3?

a. it is reduced by a factor of 3

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3 years ago

Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which wou

lyudmila [28]

Answer:

w = 1.976 rpm

Explanation:

For simulate the gravity we will use the centripetal aceleration $a_c$ , so:

$a_c = w^2r$

where w is the angular aceleration and r the radius.

We know by the question that:

r = 60.5m

$a_c$ = 2.6m/s2

So, Replacing the data, and solving for w, we get:

$2.6m/s = w^2(60.5m)$

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3 years ago

Two identical loudspeakers 2.30 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standi

Dimas [21]

Answer:

Abby is standing (4.5^2 + 2.3^2)^1/2 from the far speaker

D2 = 5.05 m from the far speaker

The difference in distances from the speakers is

5.05 - 4.5 = .55 m (Let y be wavelength, lambda)

n y = 4.5

(n + 1) y = 5.05 for the speakers to be in phase at smallest wavelength

y = .55 m subtracting equations

f = v / y = 340 / .55 = 618 / sec should be the smallest frequency

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3 years ago

Consider two laboratory carts of different masses but identical kinetic energies and the three following statements. I. The one

kolbaska11 [484]

Answer:

d) I and III only.

Explanation:

Let be $m_{1}$ and $m_{2}$ the masses of the two laboratory carts and let suppose that $m_{1} > m_{2}$ . The expressions for each kinetic energy are, respectively:

$K = \frac{1}{2}\cdot m_{1}\cdot v_{1}^{2}$ and $K = \frac{1}{2}\cdot m_{2}\cdot v_{2}^{2}$ .

After some algebraic manipulation, the following relation is constructed:

$\frac{m_{1}}{m_{2}} = \left(\frac{v_{2}}{v_{1}}\right)^{2}$

Since $\frac{m_{1}}{m_{2}} > 1$ , then $\frac{v_{2}}{v_{1}} > 1$ . That is to say, $v_{1} < v_{2}$ .

The expressions for each linear momentum are, respectively:

$p_{1} = \frac{2\cdot K}{v_{1}} = m_{1}\cdot v_{1}$ and $p_{2} = \frac{2\cdot K}{v_{2}} = m_{2}\cdot v_{2}$

Since $v_{1} < v_{2}$ , then $p_{1} > p_{2}$ . Which proves that statement I is true.

According to the Impulse Theorem, the impulse needed by cart I is greater than impulse needed by cart II, which proves that statement II is false.

According to the Work-Energy Theorem, both carts need the same amount of work to stop them. Which proves that statement III is true.

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3 years ago