in the international space station which orbits earth, astronauts experience apparent weightlessness because

On his honeymoon, James Joule attempted to explore the relationships between various forms of energy by measuring the rise of te

Marizza181 [45]

Answer:

The maximum temperature rise = 0.047 °C

Explanation:

Potential Energy, P = mgh

Energy transfered, Q=mcΔT

Potential energy = Energy transfered

mgh = mcΔT

gh = cΔT

ΔT = gh/c

ΔT = (9.81 * 20) / 4186

ΔT = 0.047 °C

8 0

3 years ago

Ann walked 1.5 miles south to her house in 0.5 hours. what is Ann's speed? what is Ann's velocity

ipn [44]

To calculate the speed and velocity of the Ann`s we use the formula,

$v= \frac{d}{t}$

Here, d is distance and t is time and v if we take it with direction then it is called velocity and if we take it without the direction then it is called speed.

Given $d=1.5 miles$ and $t=0.5 hours$ .

Substituting these values in above equation we get

$v=\frac{1.5mi }{0.5 h} = 3 mi/h$

As Ann walked towards south direction therefore, Ann`s velocity is 3 mi/h south and her speed is 3 mi/h .

7 0

3 years ago

QUESTION 9

Alik [6]

Answer:

The answer is option a.

Hope this helps

4 0

3 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

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