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3 years ago

Sunidhi made a study chart about changes in states of

Physics

1 answer:

ivann1987 [24]3 years ago

6 0

Answer:

freezing melting sublimating x y

x and

Explanation:

To drop a table of states of matter the best headings must be

table name: matter state

freezing melting sublimation x y

This order goes from the solid state to the gaseous state, which is obtained by introducing heat into the system

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Normally light waves move in all different directions. When light becomes P________, all of the electric fields in the waves mov

Aloiza [94]

Answer:im so sorry i cant find anything either ask your teacher for some help is the best thing i can do

4 0

3 years ago

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A car travels 40 miles in 30 minutes.

lukranit [14]

Answer:

(a)Average velocity ,v =128.74 Km/hr

(b)Kinetic Energy , K=958546.875 Joule

(c)Distance, s=268.8m

(d)Acceleration, a= - 2.38 $m/s^2$

<u>Explanation</u>:

<u>Given</u>:

Distance travelled = 40 miles

Time taken = 30 minutes.

(A) The average velocity in kilometres/hour

Converting 40 miles into km ,

we know that,

1 mile = 1.60934

40 miles = 40 x 1.60934

so 40 miles = 64.3738 Km

similarly converting 30 minutes into hours

1 minute = $\frac{1}{60}hours$

30 minute = $\frac{30}{60}hours$

30 minute = $\frac{1}{2}hours$

Now

Average velocity = $\frac{Speed}{time}$

Substituting Values,

Average velocity = $\frac{64.3738}{\frac[1}{2}}$

Average velocity = $64.3738 \times 2$

Average velocity =128.74 Km/hr

(B) If the car weighs 1.5 tons, what is its If the car weighs 1.5 tons, what is its kinetic energy in joules (Note: you will need to convert your velocity to m/s)? in joules (Note: you will need to convert your velocity to m/s)?

Converting 1.5 tons into kg we get

1 ton = 1000 kg

so 1.5 ton =1500 kg

converting velocity to m/s

$128.74 \times \frac{5}{18}$

=>35.75 m/s----------------------------------------------------------(1)

kinetic energy K= $\frac{1}{2}mv^2$

Substituting the values,

K= $\frac{1}{2}1500(35.75)^2$

K= $\frac{1}{2}1500(1278.06)$

K= $\frac{1500 \times (1278.06)}{2}$

K= $\frac{1917093.75}{2}$

K=958546.875 Joule---------------------------------------------(2)

(c)When the driver applies the brake, it takes 15 seconds to stop. How far does the car travel (in meters) while stopping

Lets use Distance formula,

$S= ut+\frac{1}{2}at^2$

Substituting the known values,

$s= ut+\frac{1}{2}at^2$

$s= (37.75)(15)+\frac{1}{2}a(15)^2$

$s=566.25+\frac{1}{2}a(225)$

$s=566.25+\frac{(225a)}{2}$ -------------------------------------(3)

(D) What is the average acceleration of the car (in m/s2) during braking?

Using the formula

v=u +at

re arranging the formula we get,

$a = \frac{v - u}{t}$

Substituting the values

$a = \frac{0 - 35.75}{15}$

$a = \frac{- 35.75}{15}$

a= - 2.38 $m/s^2$ ----------------------------------------(4)

Now substituting 4 in 3 we get

$s=566.25+\frac{(225( - 2.38)}{2}$

$s=566.25+\frac{-535.5}{2}$

$s=536.25-267.75$

s=268.8m--------------------------------------------------------------(5)

4 0

3 years ago

What is the unit of power

marta [7]

Answer:

The unit of power is watt.

Explanation:

8 0

3 years ago

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Problem Page Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 28. g o

sergejj [24]

Answer : The maximum mass of carbon dioxide produced by the chemical reaction can be, 82.3 grams.

Solution : Given,

Mass of $C_2H_6$ = 28 g

Mass of $O_2$ = 190 g

Molar mass of $O_2$ = 32 g/mole

Molar mass of $C_2H_6$ = 30 g/mole

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the moles of $C_2H_6$ and $O_2$ .

$\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{28g}{30g/mole}=0.933moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{190g}{32g/mole}=5.94moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O$

From the balanced reaction we conclude that

As, 2 mole of $C_2H_6$ react with 7 mole of $O_2$

So, 0.933 moles of $C_2H_6$ react with $\frac{7}{2}\times 0.933=3.26$ moles of $O_2$

From this we conclude that, $O_2$ is an excess reagent because the given moles are greater than the required moles and $C_2H_6$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CO_2$

From the reaction, we conclude that

As, 2 mole of $C_2H_6$ react to give 4 mole of $CO_2$

So, 0.933 moles of $C_2H_6$ react to give $\frac{4}{2}\times 0.933=1.87$ moles of $CO_2$