A 12 V battery is connected to a circuit with a 4 resistor. What is the current through the resistor?

5.00 kg of liquid water is heated to 100.0 °C in a closed system. At this temperature, the density of liquid water is 958 kg/m3.

user100 [1]

Answer:

$1.04\times 10^7\ J.$

Explanation:

In the question given :

Pressure is constant

Therefore, Work done, $W=P\times\Delta V$

Pressure, P=1.01 × 105 Pa.

Final volume, $V_f=8.50\ m^3.$

Initial volume, $V_i=\dfrac{Mass}{density}=\dfrac{5}{958}=5.22\times10^-3\ m^3.$

Therefore, W=8.58\times 10^{5}\ J.

Also, Heat Given, $Q=m\times L=5\times 2.26\times 10^{6}\ J=1.13\times 10^7\ J.$

Also, according to First law of thermodynamics:

$\Delta U=Q-W=(1.13\times 10^7)-(8.58\times 10^5)=1.04\times 10^7\ J.$

Hence, this is the required solution.

8 0

3 years ago

An automobile rounds a curve of radius 50.0 m on a flat road.

bixtya [17]

Answer:

14m/s

Explanation:

Given parameters:

Radius of the curve = 50m

Centripetal acceleration = 3.92m/s²

Unknown:

Speed needed to keep the car on the curve = ?

Solution:

The centripetal acceleration is the inwardly directly acceleration needed to keep a body along a curved path.

It is given as;

a = $\frac{v^{2} }{r}$

a is the centripetal acceleration

v is the speed

r is the radius

Now insert the parameters and find v;

v² = ar

v² = 3.92 x 50 = 196

v = √196 = 14m/s

6 0

3 years ago

Complete the sentences to describe the difference between speed and velocity.

xeze [42]

Answer:

velocity =displacement/time

and speed =distance/time

6 0

3 years ago

While standing atop a building 49.6 m tall, you spot a friend standing on a street corner. Using a protractor and a dangling plu

olga nikolaevna [1]

Answer:

75degree don't forget wind and gravity force pulling down

6 0

3 years ago

A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate

Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is $\bf{2.81 \times 10^{4}~n~C^{-1}}$ .

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and ' $\epsilon_{0}$ ' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

$C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

$C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF$

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

$&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm$

Part(c):

If we apply Gauss' law of electrostatics, then

$&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}$

3 0

3 years ago