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1 year ago

which of the following instruments measures air pressure? group of answer choices barometer hygrometer radiometer thermometer

1 answer:

4 0

The tool used to gauge atmospheric pressure is a barometer. Concerning the device used to gauge air pressure, we have been questioned.

Mercury is used in barometers. A reservoir of mercury is compressed by atmospheric pressure, and depending on the force of the air, the mercury is propelled up into the tube to a specific height. A straightforward barometer consists of a long glass tube that is filled with mercury and is closed at one end and open at the other. As most cases, the air pressure is expressed in mm Hg.
By delivering quick, precise information on a range of metrics, Radiometer's products and solutions aid in the diagnosis and treatment of critically ill hospital patients.

A hygrometer is a device that gauges the moisture content of soil, air, or restricted spaces. Measurements of other quantities, such as temperature, pressure, mass, or a mechanical or electrical change in a substance as moisture is absorbed, are typically used as the basis for humidity measurement systems.
Voltmeters are used to measure voltages, while thermometers measure temperature.

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A person wants to determine the spring constant of an exercise stretch cord. He pulls the cord with a force probe that exerts a

Burka [1]

Answer:

350 N/m

Explanation:

If we are assuming the stretch does not exceed the elastic range of the material, then by Hooke's law the spring constant of the cord is simply the ratio between the force 70N acting on the cord to stretch 20cm or 0.2m

k = 70 / 0.2 = 350 N/m

The spring constant is 350 N/m

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3 years ago

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Which of the following type of personal debts is usually considered wise and justified? A. credit card purchases of luxuries tha

NeTakaya

Answer:

C. Money to pay off other debts and interest

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3 years ago

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a car traveling at a velocity of 2 m/s undergoes an acceleration of 4.5 m/s^2 over a distance of 340 m. How fast will it be goin

ra1l [238]

Vi = 2m/s
a= 4.5 m/s
d= 340 m
vf= ?

use this equation ... vf^2=vi<span>^2+2ad

you should get vf = 55.3
hope this helps </span>

3 0

3 years ago

Two identical asteroids travel side by side while touching one another. If the asteroids are composed of homogeneous pure iron a

victus00 [196]

Answer:

diameter = 21.81 ft

Explanation:

The gravitational force equation is:

$F=\frac{GMm}{R^{2} }$

Where:

F => Gravitational force or force of attraction between two masses
M => Mass of asteroid 1
m => Mass of asteroid 2
R => Distance between asteroids 1 and 2 (from center of gravity)

We also know that the asteroids are identical so their masses are identical:

Since R is the distance between centers of the two asteroids and their diameters are identical (see attachment), we can conclude that:

R=d=2r

We don´t know the mass of the asteroids but we know they are composed of pure iron, so we can relate their masses to their density:

m=ρV

This is going to be helpful because the volume of a sphere is:

$\frac{4}{3}\pi r^{3}$

And know we can write our original force of gravity equation in terms of the radius of the asteroids:

$F=\frac{GMm}{R^{2} } =\frac{Gmm}{(2r)^{2} } =\frac{Gm^{2} }{4r^{2} }$
$F=\frac{G ( \frac{4}{3}\pi r^{3}ρ)^{2} }{4r^{2} }$
$F= \frac{G(16)\pi ^{2} r^{6} ρ^{2}}{(9)(4)r^{2} } =\frac{G(16)\pi ^{2} r^{4}ρ^{2} }{36}$

Now let´s plug in the values we know:

$F = 1 lb$ mutual gravitational attraction force
$G = 6.67(10)^{-11}$ gravitational constant
$ρ_{iron} =491.5 \frac{lb}{ft^{3} }$

$1= \frac{6.67(10)^{-11} \pi ^{2} r^{4} (491.5)^{2}}{36}$

Solve for r and multiply by 2 because 2r = diameter

$d=2\sqrt[4]{\frac{1}{7.07(10)^{-5} } }$

Result is d = 21.81 Feet

6 0

3 years ago

How do you ask someone out?

vazorg [7]

talk, make conversation. text isnt always the answer, but sadly now it is. text them or better yet, call them. you can talk and then do not just blurt it out. say it calmly. they say yes, congrats, but if they say no, do not have a mental breakdown. it might ruin a friendship.

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3 years ago

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