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3 years ago

5

Read the following statements. Which statement provides the correct definition of weather? Select your answer from the options b

elow. * 1 point
Atmospheric conditions at a specific time in a specific area
Atmospheric conditions over a period of time in a specific area
Average atmospheric conditions over a period of time in a region
Average atmospheric conditions across the world

1 answer:

babymother [125]3 years ago

6 0

Answer:

In one day, a store sells 14 pairs of jeans. The 14 jeans represent 20% of the total number of items sold that day. How many items did the store sell in one day? Explain or show how you got your answer.

plz help with the qsn

Explanation:

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What happened to the balloon when it was placed on the bottle with the baking soda and vinegar and why.

user100 [1]

Answer:

The ballon would be inflated. The reason is that the sodium bicarbonate in baking soda reacts with acetic acid in vinegar to produce gas.

Explanation:

The main component of baking soda is sodium bicarbonate, ${\rm Na_{2}CO_{3}}$ .

Vinegar is mostly a solution of acetic acid ${\rm CH_{3}COOH}$ in water.

Acids such as acetic acid react with carbonate salts. One of the products of such reactions is carbon dioxide ${\rm CO_{2}}$ , a gas.

In this question, when the acetic acid in vinegar reacts with sodium bicarbonate in the baking soda, the following reaction would occur:

$\begin{aligned}& {\rm Na_{2}CO_{3}}\, (aq) + 2\, {\rm CH_{3}COOH}\, (aq) \\ &\to 2\, {\rm CH_{3}COONa}\, (aq) + {\rm CO_{2}}\, (g)\end{aligned}$ .

The ${\rm CO_{2}}$ produced would then inflate the ballon placed on the opening of the bottle.

5 0

2 years ago

Steam is leaving a 4-L pressure cooker whose operating pressure is 150 kPa. It is observed that the amount of liquid in the cook

Anna71 [15]

Answer:

The mass flow rate is 2.37*10^-4kg/s

The exit velocity is 34.3m/s

The total flow of energy is 0.583 KJ/KgThe rate at which energy leave the cooker is 0.638KW

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7 0

2 years ago

How do you find the scientific period and frequency

Hunter-Best [27]

Period is the inverse of freguency (1/f) and freqquency is the inverse of period (1/p)

5 0

3 years ago

While helping an astronomy professor, you discover a binary star system in which the two stars are in circular orbits about the

Grace [21]

Complete Question

While helping an astronomy professor, you discover a binary star system in which the two stars are in circular orbits about the system's center of mass. From their color and brightness, you determine that each star has the same mass as our Sun. The orbital period of the pair is 29.3 days , based on the oscillation of brightness observed as one star occludes (hides) the other. From this information you are able to ascertain the distance between the stars.

Calculate the distance between the stars.

Express your answer to two significant digits and include the appropriate units.

Answer:

The distance between the stars is $d =3.50*10^{30}m$

Explanation:

From this question we are told that

The orbital period is $29.3 \ days = 29.3 *24 *60* 60 = 210960\ sec$

The mass of the stars are = Mass of sun $= 1.99*10^{30}kg$

For the two stars to keep on rotating on the circular orbit, the gravitational force must equal to the centripetal force and this can be mathematically represented as

$\frac{Gm^2}{d^2} = \frac{mv^2}{r}$

Where r is the radius of the circular orbit

G is the gravitational constant $=6.67 *10 ^{-11}$

d is the linear distance between the two stars = 2 r

This because during their oscillation around the circular orbit one usually hides the other

$v = \frac{2 \pi r }{T}$

Now substituting values into the above relation

$\frac{Gm^2}{(2r)^2} = \frac{m[\frac{2 \pi r}{T} ]^2}{r}$

$\frac{Gm^2}{(2r)^2} = \frac{m[\frac{4 \pi^2 r^2 }{T} ]}{r}$

$r^3 = [\frac{GmT^2}{16 \pi^2} ]$

Substituting values we have

$r = [\frac{(6.67*10^{-11})(1.99*10^{30})(2531520)^2}{16(3.142)^2} ]^{\frac{1}{3} }$

$=\sqrt[3]{5.38529*10^{30}}$

$=1.75*10^{30}m$

Now d = 2r

=> d $=2 *1.75*10^{30}m$

$=3.50 *10^30m$

6 0

4 years ago

what is the kinetic energy of a 1 kilogram ball is thrown into the air with an initial velocity of 3 m/sec

s344n2d4d5 [400]

Data:
m (mass) = 1 Kg
s (speed) = 3 m/s
Kinetic energy = ? (Joule)

Formula (Kinetic energy)
$E_{k} = \frac{m*s^2}{2}$

Solving:
$E_{k} = \frac{m*s^2}{2}$
$E_{k} = \frac{1*3^2}{2}$
$E_{k} = \frac{1*9}{2}$
$E_{k} = \frac{9}{2}$
$\boxed{\boxed{E_{k} = 4.5\:J}}\end{array}}\qquad\quad\checkmark$

3 0

4 years ago