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3 years ago

3. If I have a block of solid gold and I continually add energy to it, will it continually increase in

1 answer:

7 0

The gold was heated at rates too fast for the electrons absorbing the light energy to collide with surrounding atoms and lose energy.

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A Ray of light in air is incident on an air to glass boundary at an angle of 30. degrees with the normal. If the index of refrac

elena55 [62]

Answer:

Explanation:

Snell's law states:

n₁ sin θ₁ = n₂ sin θ₂

where n is the index of refraction and θ is the angle of incidence (relative to the normal).

The index of refraction of air is approximately 1. So:

1 sin 30° = 1.52 sin θ

θ ≈ 19°

5 0

3 years ago

The sound level at a distance of 1.48 m from a source is 120 dB. At what distance will the sound level be 70.7 dB?

Tju [1.3M]

Answer:

The second distance of the sound from the source is 431.78 m..

Explanation:

Given;

first distance of the sound from the source, r₁ = 1.48 m

first sound intensity level, I₁ = 120 dB

second sound intensity level, I₂ = 70.7 dB

second distance of the sound from the source, r₂ = ?

The intensity of sound in W/m² is given as;

$dB = 10 Log[\frac{I}{I_o} ]\\\\For \ 120 dB\\\\120 = 10Log[\frac{I}{1*10^{-12}}]\\\\12 = Log[\frac{I}{1*10^{-12}}]\\\\10^{12} = \frac{I}{1*10^{-12}}\\\\I = 10^{12} \ \times \ 10^{-12}\\\\I = 1 \ W/m^2$

$For \ 70.7 dB\\\\70.7 = 10Log[\frac{I}{1*10^{-12}}]\\\\7.07 = Log[\frac{I}{1*10^{-12}}]\\\\10^{7.07} = \frac{I}{1*10^{-12}}\\\\I = 10^{7.07} \ \times \ 10^{-12}\\\\I = 1 \times \ 10^{-4.93} \ W/m^2$

The second distance, r₂, can be determined from sound intensity formula given as;

$I = \frac{P}{A}\\\\I = \frac{P}{\pi r^2}\\\\Ir^2 = \frac{P}{\pi }\\\\I_1r_1^2 = I_2r_2^2\\\\r_2^2 = \frac{I_1r_1^2}{I_2} \\\\r_2 = \sqrt{\frac{I_1r_1^2}{I_2}} \\\\r_2 = \sqrt{\frac{(1)(1.48^2)}{(1 \times \ 10^{-4.93})}}\\\\r_2 = 431.78 \ m$

Therefore, the second distance of the sound from the source is 431.78 m.

7 0

3 years ago

(Thanks in advance)

Dmitrij [34]

I would say it B) in the same direction feel free to ask me more question

5 0

3 years ago

An object that is negatively charged could contain only electrons with no accompanying protons. O True False

marta [7]

Answer:

community services high school graduates jevs human service providers continue

6 0

3 years ago

A 10 kg ball strikes a wall with a velocity of 3 m/s to the left. The ball bounces off with a velocity of 3 m/s to the right. If

lord [1]

Answer:

The force is 272.73 newtons

Explanation:

We're going to use impulse-momentum theorem that states impulse is the change on the linear momentum this is:

$\overrightarrow{J}=\overrightarrow{p}_{f}-\overrightarrow{p}_{i}$ (1)

Impulse is also defined as average force times the time the force is applied:

$\overrightarrow{J}=\overrightarrow{F}_{avg}(\varDelta t)$ (2)

By (2) on (1):

$\overrightarrow{F}_{avg}(\varDelta t)= \overrightarrow{p}_{f}-\overrightarrow{p}_{i}$

solving for $\overrightarrow{F}_{avg}$ :

$\overrightarrow{F}_{avg}=\frac{\overrightarrow{p}_{f}-\overrightarrow{p}_{i}}{\varDelta t}$ (3)

We already know Δt is equal to 0.22 s, all we should do now is to find $\overrightarrow{p}_{f}-\overrightarrow{p}_{i}$ and put on (3) ( $\overrightarrow{p_{i}}$ the initial momentum and $\overrightarrow{p_{f}}$ the final momentum). Linear momentum is defined as $\overrightarrow{p}=m\overrightarrow{v}$ , using that on (3):

$\varDelta\overrightarrow{p}=m \overrightarrow{v_{f}}-m \overrightarrow{v_{i}}$ (4)

Velocity (v) are vectors so direction matters, if positive direction is the right direction and negative direction left $\overrightarrow{v_{i}}=+3\, \frac{m}{s}$ and $\overrightarrow{v_{f}}=-3\, \frac{m}{s}$ so (4) becomes:

$\varDelta\overrightarrow{p}=m(-3\frac{m}{s}- (+3\frac{m}{s}))=-(10kg)(6\frac{m}{s})$

$\varDelta\overrightarrow{p}=-60\, \frac{mkg}{s}$ (5)

Using (5) on (3):

$\overrightarrow{F}_{avg}=\frac{-60\, \frac{mkg}{s}}{0.22s}$

$F_{avg}=272.73N$

8 0

4 years ago