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3 years ago

When can liquid water remain above 100 C?

1 answer:

3 0

Well,

Typically, a substance is under 1 atmosphere of pressure, or 1 atm. More than 1 atm means there is more pressure than that which the earth's atmosphere exerts on an object near the surface of the earth.

If a liquid is given enough energy, and the atmospheric pressure remains constant, the liquid will turn into a gas. In the case of water, it will turn into water vapor at 100 °C. However, if you increase the pressure to greater than 1 atm, the water will be pushed together, keeping it liquid.

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- What are (a) the x component, (b) the y component, and (c) the z component of r = a - b +c if a= 7.8 + 6.6 - 7.1 , b= -2.9 + 7

frez [133]

Answer:

Explanation:

a= 7.8i + 6.6j - 7.1k

b= -2.9 i+ 7.4 j+ 3.9k , and

c = 7.6i + 8.8j + 2.2k

r = a - b +c

=7.8i + 6.6j - 7.1k - ( -2.9i + 7.4j+ 3.9k )+ ( 7.6i + 8.8j + 2.2k)

= 7.8i + 6.6j - 7.1k +2.9i - 7.4j- 3.9k )+ 7.6i + 8.8j + 2.2k

= 18.3 i +18.3 j - k

the angle between r and the positive z axis.

cosθ = 18.3 / √18.3² +18.3² +1

the angle between r and the positive z axis.

= 18.3 / 25.75

cos θ= .71

45 degree

6 0

3 years ago

Make a rough estimate of the number of quanta emitted in one second by a 100 W light bulb. Assume that the typical wavelength em

mixas84 [53]

Answer:

#_photon = 5 10²⁰ photons / s

Explanation:

For this exercise let's calculate the energy of a single quantum of energy, use Planck's law

E = h f

c= λ f

E = h c / λ

λ= 1000 nm (1 m / 109 nm) = 1000 10⁻⁹ m

Let's calculate

E₀ = 6.6310⁻³⁴ 3 10⁸/1000 10⁻⁹

E₀ = 19.89 10⁻²⁰ J

This is the energy emitted by a photon let's use a proportions rule to find the number emitted in P = 100 w

#_photon = P / E₀

#_photon = 100 / 19.89 10⁻²⁰

#_photon = 5 10²⁰ photons / s

6 0

3 years ago

Help me please forget about you and she was

ra1l [238]

The figure is showing a volume of 2.4 mL becuase it's feel 4 little segments.

Therefore, the answer is 2.4 mL.

5 0

1 year ago

A cat dozes on a stationary merry-go-round, at a radius of 5.5 m from the center of the ride. Then the operator turns on the rid

bixtya [17]

Answer:

$u_{s}=0.56$

Explanation:

For the cat to stay in place on the merry go round without sliding the magnitude of maximum static friction must be equal to magnitude of centripetal force

$F_{s.max}=\frac{mv^2}{r} \\$

Where the r is the radius of merry-go-round and v is the tangential speed

but

$F_{s.max}=u_{s}F_{N}=u_{s}mg$

So we have

$u_{s}mg=\frac{mv^2}{r}\\ u_{s}=\frac{v^2}{gr}\\ Where\\v=\frac{2\pi R}{T} \\So\\u_{s}=\frac{(\frac{2\pi R}{T} )^2}{gr} \\u_{s}=\frac{4\pi^2 r}{gT^2}$