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3 years ago

Which of these is the most likely prediction that the meteorologist would make about the weather? There is a possibility of

Physics

1 answer:

riadik2000 [5.3K]3 years ago

6 0

Answer:

B. Dust storm

Explanation:

A meteorologist would predict dust storm most likely. Before dust storm the air condition and air pressure reduces in the particular area which indicate towards a dust storm. As pressure of a region reduces the air of the neighboring area try to maintain that pressure and rush towards low pressure area and causes dust storm.

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State one use and one disadvantage of the expansion of materials when they are heated.

Reil [10]

An example of the use of expansion of materials when they are heated is in

Thermometers. Thermometer is an instrument which is used to measure the

temperature of a place.

It contains mercury which expands as temperature increases. This is used

to accurately determine the temperature of an object.

The disadvantage of expansion of materials when they are heated can be

seen in roads as there is expansion of the surface as a result of heat

exposure leads to roads being rough thereby incurring more costs for

repairs.

Read more about Heat expansion on brainly.com/question/1166774

8 0

2 years ago

In order to place a satellite into orbit, it requires enough fuel to supply the necessary mechanical energy. Into what types of

nexus9112 [7]

When a satellite is revolving into the orbit around a planet then we can say

net centripetal force on the satellite is due to gravitational attraction force of the planet, so we will have

$F_g = F_c$

$\frac{GM_pM_s}{r^2} = \frac{M_s v^2}{r}$

now we can say that kinetic energy of satellite is given as

$KE = \frac{1}{2}M_s v^2$

$KE = \frac{GM_sM_p}{2r}$

also we know that since satellite is in gravitational field of the planet so here it must have some gravitational potential energy in it

so we will have

$U = -\frac{GM_sM_p}{r}$

so we can say that energy from the fuel is converted into kinetic energy and gravitational potential energy of the satellite

6 0

3 years ago

During which moon phase do spring tides occur?

Murrr4er [49]

Answer:

The Full Moon and New Moon

Explanation:

6 0

3 years ago

What is a asteroid traveling rapidly called

SCORPION-xisa [38]

Answer:

meteor

Explanation:

A asteroid stays still and a meteor goes fast

4 0

3 years ago

Read 2 more answers

Two cylinders each contain 0.30 mol of a diatomic gas at 320 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cy

Svetllana [295]

Answer :

(a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

Explanation :

Given that,

Number of mole n = 0.30 mol

Initial temperature = 320 K

Pressure = 3.0 atm

Final pressure = 1.0 atm

We need to calculate the initial volume

Using formula of ideal gas

$P_{1}V_{1}=nRT$

$V_{1}=\dfrac{nRT}{P_{1}}$

Put the value into the formula

$V_{1}=\dfrac{0.30\times8.314\times320}{3.039\times10^{5}}$

$V_{1}=2.62\times10^{-3}\ m^3$

(a). We need to calculate the final temperature of the gas in the cylinder A

Using formula of ideal gas

In isothermally, the temperature is not change.

So, the final temperature of the gas in the cylinder A is 320 K.

(b). We need to calculate the final temperature of the gas in the cylinder B

Using formula of ideal gas

$T_{2}=T_{1}\times(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}-1}$

Put the value into the formula

$T_{2}=320\times(\dfrac{3}{1})^{\frac{1}{1.4}-1}$

$T_{2}=233.7\ K$

(c). We need to calculate the final volume of the gas in the cylinder A

Using formula of volume of the gas

$P_{1}V_{1}=P_{2}V_{2}$

$V_{2}=\dfrac{P_{1}V_{1}}{P_{2}}$

Put the value into the formula

$V_{2}=\dfrac{3\times2.62\times10^{-3}}{1}$

$V_{2}=0.00786\ m^3$

$V_{2}=7.86\times10^{-3}\ m^3$

(d). We need to calculate the final volume of the gas in the cylinder B

Using formula of volume of the gas

$V_{2}=V_{1}(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}}$

$V_{2}=2.62\times10^{-3}\times(\dfrac{3}{1})^{\frac{1}{1.4}}$

$V_{2}=0.0057\ m^3$

$V_{2}=5.7\times10^{-3}\ m^3$

Hence, (a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

6 0

3 years ago