2. Tony sets up an experiment inside his classroom. He spaces 5 plants at equal

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

3 years ago

A rectangular loop with dimensions 4.20 cm by 9.50 cm carries current I. The current in the loop produces a magnetic field at th

tiny-mole [99]

Answer:

I=1.48 A

Explanation:

Given that

B=3.1 x 10⁻5 T

b= 4.2 cm

l= 9.5 cm

The relationship for magnetic field and current given as

$B=\dfrac{2\mu _oI}{\pi}D$

Where

$D=\dfrac{\sqrt{l^2+b^2}}{lb}$

By putting the values

$D=\dfrac{\sqrt{l^2+b^2}}{lb}$

$D=\dfrac{\sqrt{0.042^2+0.095^2}}{0.042\times 0.095}$

D=26.03 m⁻¹

$B=\dfrac{2\mu _oI}{\pi}D$

$3.1\times 10^{-5}=\dfrac{2\times 4\times \pi \times 10^{-7} I}{\pi}\times 26.03$

$I=\dfrac{3.1\times 10^{-5}}{{2\times 4\times 10^{-7} }\times 26.03}$

I=1.48 A

8 0

3 years ago

Two parallel wires carry a current I in the same direction. Midway between these wires is a third wire, also parallel to the oth

Luda [366]

Answer:

Force is repulsive hence direction of force is away from wire

Explanation:

The first thing will be to draw a figure showing the condition,

Lets takeI attractive force as +ve and repulsive force as - ve and thereafter calculating net force on outer left wire due to other wires, net force comes out to be - ve which tells us that force is repulsive, hence direction of force is away from wire as shown in figure in the attachment.

4 0

3 years ago

Read 2 more answers

What relationship do you see between a star colour and temperature

Andrei [34K]

Answer:

Stars emit colors of many different wavelengths, but the wavelength of light where a star's emission is concentrated is related to the star's temperature - the hotter the star, the more blue it is; the cooler the star, the more red it is

5 0

3 years ago

Please help me find the equations guys

Alexeev081 [22]

The line at the bottom of the picture ... probably the first line on a list of choices .. is the correct equation.

4 0

2 years ago