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Identify a situation in which you would want to have a high

Andreyy89

Answer: A voltmeter must have a high resistance where as an ammeter must have a low resistance.

Explanation:

A voltmeter is a device which is connected in parallel to the component across which voltage needs to be measured. In a parallel circuit voltage drop is same at the nodes. The parallel connection must not offer easier path for current to divert from the main circuit and travel. Thus, a voltmeter must have high resistance.

On the other hand, an ammeter which is used to measure current in the circuit must have low resistance as it is connected in series. It should not offer resistance as it would reduce the actual current and measurement would be inaccurate.

7 0

3 years ago

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Which is an example of the gravitational force?

Evgesh-ka [11]

1. A basketball was thrown in the air and falls to the ground

7 0

2 years ago

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A body in simple harmonic motion has a displacement x that varies in time t according to the equation x = 5cos(π t + π/3) , wher

zhuklara [117]

Answer:

1/2 Hz

Explanation:

A simple harmonic motion has an equation in the form of

$x(t) = Acos(\omega t - \phi)$

where A is the amplitude, $\omega = 2\pi f$ is the angular frequency and $\phi$ is the initial phase.

Since our body has an equation of x = 5cos(π t + π/3) we can equate $\omega = \pi$ and solve for frequency f

$2\pi f = \pi$

f = 1/2 Hz

7 0

3 years ago

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A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent

azamat

The wavelengths of the constituent travelling waves CANNOT be 400 cm.

The given parameters:

<em>Length of the string, L = 100 cm</em>

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The wavelengths of the constituent travelling waves is calculated as follows;

$L = \frac{n \lambda}{2} \\\\n\lambda = 2L\\\\\lambda = \frac{2L}{n}$

for first mode: n = 1

$\lambda = \frac{2\times 100 \ cm}{1} \\\\\lambda = 200 \ cm$

for second mode: n = 2

$\lambda = \frac{2L}{2} = L = 100 \ cm$

For the third mode: n = 3

$\lambda = \frac{2L}{3} \\\\\lambda = \frac{2 \times 100}{3} = 67 \ cm$

For fourth mode: n = 4

$\lambda = \frac{2L}{4} \\\\\lambda = \frac{2 \times 100}{4} = 50 \ cm$

Thus, we can conclude that, the wavelengths of the constituent travelling waves CANNOT be 400 cm.

The complete question is below:

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:

A. 400 cm

B. 200 cm

C. 100 cm

D. 67 cm

E. 50 cm

Learn more about wavelengths of travelling waves here: brainly.com/question/19249186

5 0

2 years ago

Faiyez wrote the problem below

Ilya [14]

445/100 - 5/4 = 445/100 - 125/100 = 320/100 = 16/5 = 3 1/5.

6 0

3 years ago