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3 years ago

An instrument that measures air pressure using a partially evacuated chamber is called a(n ________ barometer.

1 answer:

6 0

The appropriate response is the Aneroid barometer. This kind of gauge has an incompletely cleared chamber that progressions shape, packing as barometrical weight increments and growing as weight declines.

I hope the answer will help you.

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What statement best defines energy

Rudik [331]

Energy can change forms and be transferred.

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3 years ago

A 2.0 µF capacitor is charged through a 50,000 ohm resistor. How long does it take for the capacitor to reach 90% of full charge

Nesterboy [21]

Answer:

0.23 s

Explanation:

First of all, let's find the time constant of the circuit:

$\tau=RC$

where

$R=50,000 \Omega$ is the resistance

$C=2.0\mu F=2.0\cdot 10^{-6}F$ is the capacitance

Substituting,

$\tau=(50,000 \Omega)(2.0\cdot 10^{-6}F)=0.1 s$

The charge on a charging capacitor is given by

$Q(t)=Q_0 (1-e^{-t/\tau} )$ (1)

where

$Q_0$ is the full charge

we want to find the time t at which the capacitor reaches 90% of the full charge, so the time t at which

$Q(t)=0.90 Q_0$

Substituting this into eq.(1) we find

$0.90 Q_0 = Q_0 (1-e^{-t/\tau})\\0.90=1-e^{-t/\tau}\\e^{-t/\tau}=1-0.90=0.10\\-\frac{t}{\tau}=ln(0.10)\\t=-\tau ln(0.10)=(0.1 s)ln(0.10)=0.23 s$

4 0

3 years ago

What happens to the temperature as altitude increases in the exosphere? Does it increase or decrease the higher it goes?

Whitepunk [10]

Answer:

it gets colder the higher you go

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2 years ago

Pls help if u do it 100 points

Bumek [7]

Answer:

hhuuufuhrhuuuryheu7uhhehfu

3 0

2 years ago

The parasailing system shown uses a winch to pull the rider in towards the boat, which is traveling with a constant velocity. Du

leonid [27]

Answer:

The magnitude of the force is $F_{net}= 1.837 *10^4N$

the direction is 57.98° from the horizontal plane in a counter clockwise direction

Explanation:

From the question we are told that

At t = 0 , $\theta = 20^o$

The rate at which the angle increases is $w = 2 \ ^o/s$

Converting this to revolution per second $\theta ' = 2 \ ^o/s * \frac{\pi}{180} =0.0349\ rps$

The length of the rope is defined by

$r = 125- \frac{1}{3}t^{\frac{3}{2} }$

At $\theta =30^o$ , The tension on the rope T = 18 kN

Mass of the para-sailor is $M_p = 75kg$

Looking at the question we see that we can also denote the equation by which the length is defied as an an equation that define the linear displacement

Now the derivative of displacement is velocity

$r' = -\frac{1}{3} [\frac{3}{2} ] t^{\frac{1}{2} }$

represents the velocity, again the derivative of velocity gives us acceleration

$r'' = -\frac{1}{4} t^{-\frac{1}{2} }$

Now to the time when the rope made angle of 30° with the water

generally angular velocity is mathematically represented as

$w = \frac{\Delta \theta}{\Delta t}$

Where $\theta$ is the angular displacement

Now considering the interval between $20^o \ to \ 30^o$ we have

$2 = \frac{30 -20 }{t -0}$

making t the subject

$t = \frac{10}{2}$

$= 5s$

Now at this time the displacement is

$r = 125- \frac{1}{3}(5)^{\frac{3}{2} }$

$= 121.273 m$

The linear velocity is

$r' = -\frac{1}{3} [\frac{3}{2} ] (5)^{\frac{1}{2} }$

$= -1.118 m/s$

The linear acceleration is

$r'' = -\frac{1}{4} (5)^{-\frac{1}{2} }$

$= -0.112m/s^2$

Generally radial acceleration is mathematically represented by

$\alpha _R = r'' -r \theta'^2$

$= -0.112 - (121.273)[0.0349]^2$

$= 0.271 m/s^2$

Generally angular acceleration is mathematically represented by

$\alpha_t = r \theta'' + 2 r' \theta '$

Now $\theta '' = \frac{d (0.0349)}{dt} = 0$

$\alpha _t = 121.273 * 0 + 2 * (-1.118)(0.0349)$

$= -0.07805 m/s^2$

The net resultant acceleration is mathematically represented as

$a = \sqrt{\alpha_R^2 + \alpha_t^2 }$

$= \sqrt{(-0.07805)^2 +(-0.027)^2}$

$= 0.272 m/s^2$

Now the direction of the is acceleration is mathematically represented as

$tan \theta_a = \frac{\alpha_R }{\alpha_t }$

$\theta_a = tan^{-1} \frac{-0.271}{-0.07805}$

$= 73.26^o$

The force on the para-sailor along y-axis is mathematically represented as

$F_y = mg + Tsin 30^o + ma sin(90- \theta )$

$= (75 * 9.8) + (18 *10^3) sin 30 + (75 * 0.272)sin(90-73.26)$

$= 9.74*10^3 N$

The force on the para-sailor along x-axis is mathematically represented as

$F_x = mg + Tcos 30^o + ma cos(90- \theta )$

$= (75 * 9.8) + (18 *10^3) cos 30 + (75 * 0.272)cos(90-73.26)$

$= 1.557 *10^4 N$

The net resultant force is mathematically evaluated as

$F_{net} = \sqrt{F_x^2 + F_y^2}$

$=\sqrt{(1.557 *10^4)^2 + (9.74*10^3)^2}$

$F_{net}= 1.837 *10^4N$

The direction of the force is

$tan \theta_f = \frac{F_y}{F_x}$

$\theta_f = tan^{-1} [\frac{1.557*10^4}{9.74*10^3} ]$

$= tan^{-1} (1.599)$

$= 57.98^o$

7 0

3 years ago