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Dmitrij [34]
3 years ago
6

What is the equivalent resistance of the circuit?

Physics
2 answers:
bagirrra123 [75]3 years ago
7 0

Answer:

60

Explanation:

A  P   E  X

Alona [7]3 years ago
4 0
The correct answer is option B. i.e. 60 ohm.
The diagram is attached.

Since all the resistances are attached in series.
Hence, the total resistance will be the sum of all resistances.

R_total = R_1+R_2+ R_3
= 10+20+30
= 60 ohms.

Therefore, option c is correct answer.

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Sodium Hydroxide is highly Basic
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The rate at which light energy is radiated from a source is measured in which of the following units?
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The correct option is D.
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El tubo de entrada que suministra presión de aire para operar un gato hidráulico tiene 2 cm de diámetro. El pistón de salida es
Bess [88]

Answer:

La presión neumática para levantar un automóvil de 17,640 newtons es 220,500 pascales.

Explanation:

Asumiendo que la presión (P), medida en pascales, tiene una distribución uniforme sobre la superficie del pistón, se calcula a partir de la siguiente expresion:

P = \frac{F}{A}

Donde:

F - Fuerza motriz, medida en newtons.

A - Área del pistón, medida en metros cuadrados.

La fuerza motriz es equivalente al peso del automóvil. El área del pistón (A), medido en metros cuadrados, es determinado por:

A=\frac{\pi}{4}\cdot D^{2}

Donde D es el diámetro del pistón, medido en metros.

Si D = 0.32\,m y F =17,640\,N, entonces la presión neumática es:

A = \frac{\pi}{4}\cdot (0.32\,m)^{2}

A \approx 0.080\,m^{2}

P = \frac{17,640\,N}{0.080\,m^{2}}

P = 220,500\,Pa

La presión neumática para levantar un automóvil de 17,640 newtons es 220,500 pascales.

8 0
3 years ago
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5 0
3 years ago
"What is the magnifying power of an astronomical telescope using a reflecting mirror whose radius of curvature is 5.9 m and an e
notka56 [123]

Answer:

The Magnifying power of a telescope is M = 109.26

Explanation:

Radius of curvature R = 5.9 m = 590 cm

focal length of objective f_{objective} = \frac{R}{2}

⇒ f_{objective} = \frac{590}{2}

⇒ f_{objective} = 295 cm

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Magnifying power of a telescope is given by,

M = \frac{f_{objective} }{f_{eyepiece} }

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M = 109.26

therefore the Magnifying power of a telescope is M = 109.26

4 0
3 years ago
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