Describe the responses of the human ear to sound waves coming from the

1 answer:

8 0

If the sound comes from the right side, the waves reach the right ear before the left ear. if the sound comes from the left side, the waves reach the left ear before the right ear. The difference between the phases of waves reaching both ears is detected by the ears and then interpreted by the brain

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The current in a series circuit is 19.3 A. When an additional 7.40-Ω resistor is inserted in series, the current drops to 13.4 A

Bogdan [553]

Answer:

16.8ohms

Explanation:

According to ohm's law which states that the current passing through a metallic conductor at constant temperature is directly proportional to the potential difference across its ends.

Mathematically, V = IRt where;

V is the voltage across the circuit

I is the current

R is the effective resistance

For a series connected circuit, same current but different voltage flows through the resistors.

If the initial current in a circuit is 19.3A,

V = 19.3R... (1)

When additional resistance of 7.4-Ω is added and current drops to 13.4A, our voltage in the circuit becomes;

V = 13.4(7.4+R)... (2)

Note that the initial resistance is added to the additional resistance because they are connected in series.

Equating the two value of the voltages i.e equation 1 and 2 to get the resistance in the original circuit we will have;

19.3R = 13.4(7.4+R)

19.3R = 99.16+13.4R

19.3R-13.4R = 99.16

5.9R = 99.16

R= 99.16/5.9

R = 16.8ohms

The resistance in the original circuit will be 16.8ohms

5 0

3 years ago

The volume of an object as a function of time is V(t) = At³, where A is a constant. Let L and T denote dimensions of length and

Greeley [361]

Answer:

c) L³/T³

Explanation:

If t stands for time, the units are:

(V) = L³, (t) = T

The units for the equation:

V(t) = At³

must be:

$L^{3} = \frac{L^{3} }{T^{3}} T^{3}$

7 0

2 years ago

One watt is A. equal to one kilogram per second B. the power that can be delivered by an average horse C. the same as one-foot p

Naddik [55]

Answer:

D. the proper replacement unit for one joule per second

Explanation:

When energy is divided by the time the energy was used we get power

$P=\dfrac{E}{t}\\\Rightarrow P=\dfrac{mv^2}{t}\\\Rightarrow P=\dfrac{kg(\dfrac{m^2}{s^2})}{s}\\\Rightarrow P=kg(\dfrac{m^2}{s^2})}\times \dfrac{1}{s}\\\Rightarrow P=\dfrac{kgm^2}{s^3}$