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2 years ago

You lift a stuffed toy up above the ground (over your head).

Physics

1 answer:

BigorU [14]2 years ago

7 0

Answer:

Gravitational energy

Explanation:

when over head , when it hits the floor thermal energy

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An insulating sphere is 8.00 cm in diameter and carries a 6.50 ÂµC charge uniformly distributed throughout its interior volume.

Kobotan [32]

Explanation:

(a) Formula to calculate the density is as follows.

$\rho = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}$

= $2.42 \times 10^{-2} C/m^{3}$

Now, calculate the charge as follows.

$q_{in} = \rho(\frac{4}{3} \pi r^{3})$

= $2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}$

= $10.106 \times 10^{-8}$ C

or, = 101.06 nC

(b) For r = 6.50 cm, the value of charge will be calculated as follows.

$q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}$

= 7.454 $\mu C$

7 0

3 years ago

The four giant gas planets that are made primarily of lightweight elements are

Hoochie [10]

D) although recent research (1990's) has shown Uranus and Neptune to be ice giants and have heavier substances.

8 0

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:

Given:

$T_1$ = 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 = $1.01325\times 10^5\ Pa$

Assume:

$P_2$ = 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

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