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3 years ago

Which sound wave features are being described?

Physics

2 answers:

densk [106]3 years ago

8 0

<h2>Answer:</h2>

Number of wavelengths

----------------------------------- : ⇒ Frequency

Time

Distance traveled

-------------------------- : ⇒ Speed

Time

Andre45 [30]3 years ago

6 0

Answer: Sound wave can be described by five characteristics Wavelength , amplitude, time period, frequency and velocity or speed

Explanation:

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Wherever a river loses energy

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3 years ago

9.The force of gravity between two asteroids is 10,000 newtons (N).

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Answer:

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2 years ago

Cho mạch như hình vẽ E

labwork [276]

Answer:

Explanation:

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2 years ago

A 1500-W electric heater is plugged into the outlet of a 120-V circuit that has a 20-A circuit breaker. You plug an electric hai

Korolek [52]

Answer:

b) 900 W

Explanation:

The breaker trips when the current is equal to 20 A. The power (P) is the ddp (V) multiplied by the current. So, for the electric heater, the current is:

P = V*i

1500 = 120*i

i = 12.5 A

So, to become in 20 A, it's needed 7.5 A, which must come from the hairdryer. Its power must be:

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8 0

3 years ago

An automobile battery has an emf of 12.6 V and an internal resistance of 0.0600 . The headlights together present equivalent res

murzikaleks [220]

Answer:

(a) $V=11.86\ V$

(b) $V=9.76\ V$

Explanation:

<u>Electric Circuits</u>

Suppose we have a resistive-only electric circuit. The relation between the current I and the voltage V in a resistance R is given by the Ohm's law:

$V=R.I$

(a) The electromagnetic force of the battery is $\varepsilon =12.6\ V$ and its internal resistance is $R_i=0.06\ \Omega$ . Knowing the equivalent resistance of the headlights is $R_e=5.2\ \Omega$ , we can compute the current of the circuit by using the Kirchhoffs Voltage Law or KVL:

$\varepsilon=i.R_i+i.R_e=i.(R_i+R_e)$

Solving for i

$\displaystyle i=\frac{\varepsilon}{ R_i+R_e}=\frac{12}{0.06+5.2}=2.28\ A$

i=2.28\ A

The potential difference across the headlight bulbs is

$V=\varepsilon -i.R_i=12\ V-2.28\ A\cdot 0.06\ \Omega=11.86\ V$

$V=11.86\ V$

(b) If the starter motor is operated, taking an additional 35 Amp from the battery, then the total load current is 2.28 A + 35 A = 37.28 A. Thus the output voltage of the battery, that is the voltage that the bulbs have is

$V=\varepsilon -i.R_i=12\ V-37.28\ A\cdot 0.06\ \Omega=9.76\ V$

5 0

3 years ago