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2 years ago

8

A ball is dropped from the rest from a high window of a tall building and called for 4 seconds. Neglecting air resistance, the f

inal velocity of the ball is ____ m/s. (g=-9.8m/s^2)

1 answer:

diamong [38]2 years ago

3 0

Answer:

-39.2m/s

Explanation:

Given that :

t = 4secs

g = -9.8m/s^2

v = ?

u = 0m/s ( since it was at rest )

V = u +at............. 1

Where v is the final velocity

a = -g = -9.8m/s^2 since the ball was dropped from a height which will eventually make it move against gravity

t = 4secs

Substitute the values into 1

v = 0 - 9.8×4

v = -39.2m/s

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Dry air is primarily composed of nitrogen. In a classroom demonstration, a physics instructor pours 3.6 L of liquid nitrogen int

neonofarm [45]

Answer:

The value is $V_n = 2.2498 \ m^3$

Explanation:

From the question we are told that

The volume of liquid nitrogen is $V_n = 3.6 \ L= 3.6 *10^{-3} \ m^3$

The density of nitrogen at gaseous form is $\rho_n = 1.2929 \ kg/m^3$ = The dry air at sea level

Generally the density of nitrogen at liquid form is

$\rho _l = 808 \ kg/m^3$

And this is mathematically represented as

$\rho_l = \frac{m}{V_l }$

=> $m = \rho_l * V_l$

Now the density of gaseous nitrogen is

$\rho_n = \frac{m}{V_n }$

=> $m = \rho_n * V_n$

Given that the mass is constant

$\rho_n * V_n = \rho_l * V_l$

$1.2929* V_n = 808 * 3.6*10^{-3}$

=> $V_n = 2.2498 \ m^3$

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2 years ago

The air in a car tire la compressed when the car rolls over a rock. If the air

stealth61 [152]

Answer:

the signs of heat and work are; -Q and -W

Explanation:

The first law of thermodynamics is given by; ΔU = Q − W

where;

ΔU is the change in internal energy of a system,

Q is the net heat transfer (the sum of all heat transfer into and out of the system)

W is the net work done (the sum of all work done on or by the system).

Now, The system in this case is the tire and since the air gets warmer, heat must have left the system. Therefore Q is negative (-Q).

Since work is done by the system, W remains negative.

Thus, the signs of heat and work are; -Q and - W

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2 years ago

Sports managers have the potential to earn more than a million dollars per year.<br> True<br> false

andriy [413]

Answer:

true

Explanation:

hope this helped!

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2 years ago

Which is the truest statement? Standing is health-enhancing because it uses energy. Jogging is not a good way tp enhance your he

Setler [38]

A. <span>Standing is health-enhancing because it uses energy.

Because standing can burn a lot of calories.</span>

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2 years ago

An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com

solong [7]

Answer:

(a) $3.81\times 10^5\ Pa$

(b) $4.19\times 1065\ Pa$

Explanation:

<u>Given:</u>

$T_1$ = The first temperature of air inside the tire = $10^\circ C =(273+10)\ K =283\ K$

$T_2$ = The second temperature of air inside the tire = $46^\circ C =(273+46)\ K= 319\ K$

$T_3$ = The third temperature of air inside the tire = $85^\circ C =(273+85)\ K=358 \ K$

$V_1$ = The first volume of air inside the tire

$V_2$ = The second volume of air inside the tire = $30\% V_1 = 0.3V_1$

$V_3$ = The third volume of air inside the tire = $2\%V_2+V_2= 102\%V_2=1.02V_2$

$P_1$ = The first pressure of air inside the tire = $1.01325\times 10^5\ Pa$

<u>Assume:</u>

$P_2$ = The second pressure of air inside the tire

$P_3$ = The third pressure of air inside the tire
n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

$PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)$

Part (a):

Using the above equation for this part of compression in the air, we have

$\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa$

Hence, the pressure in the tire after the compression is $3.81\times 10^5\ Pa$ .

Part (b):

Again using the equation for this part for the air, we have

$\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa$

Hence, the pressure in the tire after the car i driven at high speed is $4.19\times 10^5\ Pa$ .

8 0

2 years ago