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2 years ago

how does the frequency of a radio wave compare to the frequency of the vibrating electrons that produce it?

Physics

1 answer:

HACTEHA [7]2 years ago

3 0

Answer:

mass

Explanation:

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What will be the final temperature if a 4.00 g silver ring at 41.0◦C if it gives off 18.0 J of heat to the surroundings? The spe

Eduardwww [97]

Answer:

Final temperature, $T_f=21.85^{\circ}$

Explanation:

Given that,

Mass of silver ring, m = 4 g

Initial temperature, $T_i=41^{\circ}C$

Heat released, Q = -18 J (as heat is released)

Specific heat capacity of silver, $c=0.235\ J/g\ C$

To find,

Final temperature

Solution,

The expression for the specific heat is given by :

$Q=mc\Delta T$

$Q=mc(T_f-T_i)$

$T_f=\dfrac{-Q}{mc}+T_i$

$T_f=\dfrac{-18}{4\times 0.235}+41$

$T_f=21.85^{\circ}$

So, the final temperature of silver is 21.85 degrees Celsius.

5 0

3 years ago

Relationship between voltage, resistance and current in a circuit

Lena [83]

Answer:

V = I×R

where -

V = potential difference across

I = current flowing in the circuit

R = Equivalent Resistance in the circuit

8 0

2 years ago

Stars are formed in which of the following?

MAXImum [283]

Stars are formed in <u>nebulas</u>, interstellar clouds of dust and gas.

6 0

2 years ago

Two identical charged pith balls are brought together to touch each other. They are then

sergejj [24]

Answer:

-17.5 nC

Explanation:

charge A = -30 nC

charge B = -5 nC

After adding them it would be the average of the two charges because of the getting same voltage difference. so

c = (-30+(-5)) / 2 nC

c= -17.5 nC

answer is -17.5 nC

4 0

3 years ago

An open pipe of length 0.39m vibrates in the third harmonic with a frequency of 1400Hz. What is the distance from the center of

Gnoma [55]

Length of the pipe = 0.39 m

Third harmonic frequency = 1400 Hz

For the third harmonic:

Wavelength = $\frac{2L}{3}$

The center of the open pipe will host a node and the nearest anti - node from the center will be at the 0.25 × wavelength

Distance from center = 0.25 × wavelength

Distance = $0.25 x \frac{2L}{3}$

Plugging the value of the length of the pipe (L) = 0.39 m = 39 cm

Distance = $0.25 x \frac{2 \times 39}{3}$

Distance from the center to the nearest anti - node = 6.5 cm

Hence, the nearest distance to the anti - node from the center = 6.5 cm

So, option C is correct.

7 0

3 years ago