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Georgia [21]
3 years ago
13

Two light bulbs are installed in two sockets, the sockets are connected in series, and power is then applied to the combination

so that both bulbs light. If one of the bulbs is then removed from its socket, the other one will get brighter.
Physics
1 answer:
Arisa [49]3 years ago
5 0

Answer:

False

Explanation:

<em>If one of the bulbs is removed from the series, the other bulb will not come on at all.</em>

This is because the removal of one of the bulbs would interrupt the flow of current though the entire circuit.

Hence, that the other one will get brighter if one of two bulbs in a circuit is removed from its socket is not true.

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Maksim231197 [3]

Answer:

A,C

Explanation:

8 0
3 years ago
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A concert loudspeaker suspended high off the ground emits 34 W of sound power. A small microphone with a 1.0 cm2 area is 44 m fr
rjkz [21]

Answer:

<u>Part A</u>

I = 1.4 mW/m²  

<u>Part B</u>

β = 91.46 dB

Explanation:

<u>Part A</u>

Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.

For a spherical sound wave, the sound intensity is given by;

                                            I = \frac{P}{A}

                                            I = \frac{P}{4\pi r^{2}}

Where;

P is the source of power in watts (W)

I is the intensity of the sound in watt per square meter (W/m2)

r is the distance r away

Given:

P = 34 W,

A = 1.0 cm²

r = 44 m

The sound intensity at the position of the microphone is calculated to be;

                                     I = \frac{34}{4\pi (44)^{2}}

                                     I = \frac{34}{4\pi (44)^{2}}

                                     I = 0.0013975 W/m²

                                 ≈  I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²

                                     I = 1.4 mW/m²

The sound intensity at the position of the microphone is 1.4 mW/m².

<u>Part B</u>

Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value.  It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.

Sound intensity level is calculated as;  

                                    β = 10log_{10}\frac{I}{I_{0}}  dB

Where,

β is the Sound intensity level in decibels (dB)

I is the sound intensity;

I₀ is the reference sound intensity;

By pluging-in, I₀ is 1.0 × 10⁻¹² W/m²

           ∴        β = 10log_{10}\frac{1.4 * 10^{-3} W/m^{2}}{1.0 * 10^{-12} W/m^{2}}

                      β = 10log_{10} (1.4 * 10^{9})

                      β = 91.46 dB

The sound intensity level at the position of the microphone is 91.46 dB.                

4 0
3 years ago
The temperature of water in a beaker is 45°C. What does this measurement represent
andrezito [222]
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Black_prince [1.1K]
<span>
Reaction rates are affected by reactant concentrations and temperature. this is accounted for by the c</span>ollision model.

-Hope this helps.
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