All categories

2 years ago

PLEASE PLEASE HELP AND PUT A REAL ANSWER ;-;. ALSO WILL GIVE BRAINLEST!!

Physics

2 answers:

torisob [31]2 years ago

8 0

Troposphere is the right answer

uysha [10]2 years ago

3 0

Answer:

Troposphere

High-pressure areas form due to downward motion through the troposphere, the atmospheric layer where weather occurs.

You might be interested in

If the Earth rotated in the opposite direction, how would that affect the wind? *

Alik [6]

Answer:

The wind would still blow, but it would curve and spin in the opposite direction.

Explanation:

6 0

2 years ago

Read 2 more answers

How can two polarizing filters be used to show that light has some properties of a wave

DIA [1.3K]

Polarizing filters: used to show light has some properties of a wave in that way of property is that light can be thought of as traveling forward in waves, with the <span>wave. sorry if this is too late but google is good</span>

6 0

2 years ago

Three blocks are placed in contact on a horizontal frictionless surface. A constant force of magnitude F is applied to the box o

Lina20 [59]

Answer:

A) M

Explanation:

The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:

Box with mass M

$\Sigma F = F - F' = M\cdot a$

Box with mass 2M

$\Sigma F = F' - F'' = 2\cdot M \cdot a$

Box with mass 3M

$\Sigma F = F'' = 3\cdot M \cdot a$

On the third equation, acceleration can be modelled in terms of F'':

$a = \frac{F''}{3\cdot M}$

An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.

$F' = 2\cdot M \cdot a + F''$

$F' = 2\cdot M \cdot \left(\frac{F''}{3\cdot M} \right) + F''$

$F' = \frac{5}{3}\cdot F''$

Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:

$F - \frac{5}{3}\cdot F'' = M \cdot \left(\frac{F''}{3\cdot M} \right)$

$F = \frac{5}{3} \cdot F'' + \frac{1}{3}\cdot F''$

$F = 2\cdot F''$

$F'' = \frac{1}{2}\cdot F$

Afterwards, F' as function of the external force can be obtained by direct substitution:

$F' = \frac{5}{6}\cdot F$

The net forces of each block are now calculated:

Box with mass M

$M\cdot a = F - \frac{5}{6}\cdot F$

$M\cdot a = \frac{1}{6}\cdot F$

Box with mass 2M

$2\cdot M\cdot a = \frac{5}{6}\cdot F - \frac{1}{2}\cdot F$

$2\cdot M \cdot a = \frac{1}{3}\cdot F$

Box with mass 3M

$3\cdot M \cdot a = \frac{1}{2}\cdot F$

As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.

8 0

3 years ago

You push a block to the top of an inclined plane. The mechanical advantage of the plane is 5. The inclined plane is 10 meters lo

Veronika [31]

Answer: 50 m

Explanation:

4 0

3 years ago

Read 2 more answers

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 $CO_2$

Now we have to calculate the mass of $CO_2$

$\text{ Mass of }CO_2=\text{ Moles of }CO_2\times \text{ Molar mass of }CO_2$

$\text{ Mass of }CO_2=(1.87moles)\times (44g/mole)=82.3g$

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

6 0

3 years ago