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monitta
3 years ago
12

DEFINA O QUE É AGILIDADE?

Physics
2 answers:
scoray [572]3 years ago
7 0

capacidad de moverse rápida y fácilmente.

¡Espero que esto ayude! :)

faltersainse [42]3 years ago
3 0

Answer:

  • sorry

Explanation:

sorry lol i don't know

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9. Una jeringa contiene cloro gaseoso, que ocupa un volumen de 95 mL a una presión de 0,96 atm. ¿Qué presión debemos ejercer en
masha68 [24]

Answer:

2.61 atm

Ley de Boyle

Explanation:

P_1 = Presión inicial = 0.96 atm

P_2 = Presión final

V_1 = Volumen inicial = 95 mL

V_2 = Volumen final = 35 mL

En este problema usaremos la ley de Boyle.

\dfrac{P_1}{P_2}=\dfrac{V_2}{V_1}\\\Rightarrow P_2=\dfrac{P_1V_1}{V_2}\\\Rightarrow P_2=\dfrac{0.96\times 95}{35}\\\Rightarrow P_1=2.61\ \text{atm}

La presión ejercida sobre el émbolo para reducir su volumen es de 2.61 atm.

4 0
3 years ago
Why can an object still be seen when it is at absolute zero?
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6 0
3 years ago
PLS ANSWER FAST WILL GIVE BRAINLY TIMED TEST
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4 0
3 years ago
Consider an aircraft powered by a turbojet engine that has a pressure ratio of 12. The aircraft is stationary on the ground, hel
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3 0
3 years ago
A 4-79 permalloy solenoid coil needs to produce a minimum inductance of 1.1 . If the maximum allowed current is 4 , how many tur
daser333 [38]

The related concept to solve this exercise is given in the expressions that the magnetic field has both as a function of the number of loops, current and length, as well as inductance and permeability. The first expression could be given as,

The magnetic field H is given as,

H = \frac{nI}{l}

Here,

n = Number of turns of the coil

I = Current that flows in the coil

l = Length of the coil

From the above equation, the number of turns of the coil is,

n = \frac{Hl}{I}

The magnetic field is again given by,

H = \frac{B}{\mu_t}

Where the minimum inductance produced by the solenoid coil is B.

We have to obtain n, that

n = \dfrac{\frac{B}{\mu_t}l}{I}

Replacing with our values we have that,

n = \dfrac{\frac{1.1Wb/m^2 }{200000}(2m)}{4mA}

n = \dfrac{(\frac{1.1Wb/m^2 }{200000})(\frac{10^4 guass}{1Wb/m^2})(2m)}{4mA(\frac{10^{-3}A}{1mA})}

n = 27.5 \approx 28

Therefore the number of turn required is 28Truns

4 0
4 years ago
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