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Maslowich
2 years ago
10

An ammeter with a resistance of 5.0 ohm is connected in series with a 3.0V cell and a lamp rated at 300 mA, 3V. Calculate the cu

rrent that the ammeter will measure
Physics
1 answer:
Alex2 years ago
3 0

Answer:

I = 0.2 A

Explanation:

Lamp is rated at 300 mA

I_lamp = 0.3 A

Voltage is; V = 3V

Thus; Resistance is given by;

R = V/I

R = 3/0.3

R = 10 ohms

Now, since the ammeter of 5 ohms is connected in series with the lamp. Thus equivalent resistance;

R_eq = 10 + 5

R_eq = 15 ohms

Ammeter current will be;

I = V/R_eq

I = 3/15

I = 0.2 A

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weathering

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If 20 beats are produced within one second, which of the following frequencies could possibly be held by two sound waves traveli
NeTakaya

Answer:

D. 22 Hz and 42 Hz

Explanation:

  • When two waves with different frequency travelling in the same medium meet each other, they produce an interference pattern called beat.
  • <em><u>The frequency of the beat produced is equivalent to </u></em><em><u>the difference between the individual frequencies of the two waves involved.</u></em>
  • <em><u>Therefore; in this case since the frequency of the beat is 20 Hz, that is from 20 beats per second.</u></em>
  • We need to find a pair from the choices whose frequency difference is 20 Hz.
  • This happens to be choice D. 22 Hz and 42 Hz,  that is 42 Hz - 22 Hz = 20 Hz
8 0
3 years ago
Read 2 more answers
A wave travels at 295 m/s and has a wavelength of 2.50 m. What is the frequency of the wave?
posledela

Answer:

118\; \rm Hz.

Explanation:

The frequency f of a wave is equal to the number of wave cycles that go through a point on its path in unit time (where "unit time" is typically equal to one second.)

The wave in this question travels at a speed of v= 295\; \rm m\cdot s^{-1}. In other words, the wave would have traveled 295\; \rm m in each second. Consider a point on the path of this wave. If a peak was initially at that point, in one second that peak would be

How many wave cycles can fit into that 295\; \rm m? The wavelength of this wave\lambda = 2.50\; \rm m gives the length of one wave cycle. Therefore:

\displaystyle \frac{295\;\rm m}{2.50\; \rm m} = 118.

That is: there are 118 wave cycles in 295\; \rm m of this wave.

On the other hand, Because that 295\; \rm m of this wave goes through that point in each second, that 118 wave cycles will go through that point in the same amount of time. Hence, the frequency of this wave would be

Because one wave cycle per second is equivalent to one Hertz, the frequency of this wave can be written as:

f = 118\; \rm s^{-1} = 118\; \rm Hz.

The calculations above can be expressed with the formula:

\displaystyle f = \frac{v}{\lambda},

where

  • v represents the speed of this wave, and
  • \lambda represents the wavelength of this wave.

6 0
3 years ago
Gold, which has a mass of 19.32 g for each cubic centimeter of volume, is the most ductile metal and can be pressed into a thin
Dennis_Churaev [7]

Answer:

area = 5733.33  cm²

length = 5.47 ×10^{7} cm

Explanation:

Given data

density = 19.32 g/cm³

mass = 33.16 g

thickness = 3.000 µm = 3 ×10^{-4} cm

radius r = 1.000 µm = 1 ×10^{-4} cm

to find out

area of the leaf and  length of the fiber

solution

we know volume formula that is

volume = mass / density

volume = 33.16 /  19.32

volume = 1.72 cm³

we know that volume = thickness × area

so area

area = volume / thickness

area = 1.72 / 3 ×10^{-4}

area = 5733.33  cm²

and

we know volume = πr²L

so L = volume /  πr²

length = 1.72 / π(1×10^{-4})²

length = 5.47 ×10^{7} cm

3 0
3 years ago
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