A jet lands at 80.0 m/s, the pilot applying the brakes 2.0 s after landing.

A tennis ball was thrown straight up with initial velocity of 22.5 m/s. How high does the ball rise

kiruha [24]

Answer:

$h_m=25.8\ m$

Explanation:

<u>Vertical Launch Upwards</u>

In a vertical launch upwards, an object is launched vertically up without taking into consideration friction with the air.

If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:

$\displaystyle h_m=\frac{v_o^2}{2g}$

The tennis ball was thrown straight up with a speed of v0=22.5 m/s. The acceleration of gravity is g=9.81\ m/s^2, thus:

$\displaystyle h_m=\frac{22.5^2}{2\cdot 9.81}$

$\mathbf{h_m=25.8\ m}$

6 0

3 years ago

Which of the following statements is true of the strong nuclear force?

vredina [299]

Explanation:

it holds protons and neutrons together

5 0

3 years ago

Which type of drug would be useful in dilating the pupils for an examination of the retina? Which type of drug would be useful i

givi [52]

Answer:

muscarinic receptor inhibitor

Explanation:

6 0

3 years ago

Read 2 more answers

What do you call an increase in the magnitude of velocity

Vlad1618 [11]

Any change in the magnitude or direction of velocity is "acceleration".

4 0

3 years ago

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