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3 years ago

Question 1 of 10

Physics

1 answer:

Annette [7]3 years ago

7 0

Answer:

A. 2.0

Explanation:

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A lift is used to lower miners 5500m into a mine. The lift is supported by a cable made of many high strength steel strands of 6

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3 years ago

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.

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3 years ago

In Lesson 4.1, you learned that the machine used to fuel missions to Titan is not making enough liquid oxygen. We are waiting fo

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Answer: The ideas that I have is how useful the Universal Space Agency or USA, is how great the fuel missions are

Explanation:

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2 years ago

True or false Electrons are lot lower in mass than neurons

pickupchik [31]

True! they're the lowest in mass of all parts of the atom

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3 years ago

A proton enters a uniform magnetic field of strength 2 T at 300 m/s. The magnetic field is oriented perpendicular to the proton’

sattari [20]

Answer:

Magnetic force, $F=9.6\times 10^{-17}\ N$

Explanation:

It is given that,

Magnetic field, B = 2 T

Velocity of the proton, v = 300 m/s

Charge on the proton, $q=1.6\times 10^{-19}\ C$

The magnetic field is oriented perpendicular to the proton’s velocity. The magnetic force on the charged particle is given by :

$F=qvB\ sin\theta$

The magnetic field is oriented perpendicular to the proton’s velocity, $\theta=90^{\circ}$

$F=1.6\times 10^{-19}\times 300\times 2$

$F=9.6\times 10^{-17}\ N$

So, the magnitude of the force that the proton experiences while it moves through the magnetic field is $9.6\times 10^{-17}\ N$ . Hence, this is the required solution.

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3 years ago