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2 years ago

12

The electric field is measured for points at distances r from the center of a uniformly charged insulating sphere that has volum

e charge densi

1 answer:

RoseWind [281]2 years ago

4 0

For the electric field is measured for points at distances r,Electric field is mathematically given as

E=19.9*10^{-5}c/m^3

<h3>What is the Electric field?</h3>

Generally, the equation for the electric field uniform charged is mathematically given as

$E=\frac{lr}{3e0}$

Therefore

E=lr/3E0=3Ee0/r

Therefore

E=3*6*10^4*8.85*10^{-12}

E=19.9*10^{-5}c/m^3

In conclusion, E=19.9*10^{-5}c/m^3

E=19.9*10^{-5}c/m^3

Read more about Electric field

brainly.com/question/9383604

CQ

Complete question attached below

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Answer:

8. organelle

Explanation:

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3 years ago

A force in the +x -direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 7.90 kg box that is sitting on the horizontal

dsp73

Answer:

$v\approx 8.570\,\frac{m}{s}$

Explanation:

The equation of equlibrium for the box is:

$\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a$

The formula for the acceleration, given in $\frac{m}{s^{2}}$ , is:

$a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}$

Velocity can be derived from the following definition of acceleration:

$a = v\cdot \frac{dv}{dx}$

$v\, dv = a\, dx$

$\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx$

$\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg} \int\limits^{17\,m}_{0\,m}\, dx - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx$

$\frac{1}{2}\cdot v^{2} = (2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}$

$v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}] }$

The speed after the box has travelled 17 meters is:

$v\approx 8.570\,\frac{m}{s}$

3 0

3 years ago

An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:

Burka [1]

Heat added to the gas = Q = 743 Joules

Work done on the gas = W = -743 Joules

$\texttt{ }$

<h3>Further explanation</h3>

The Ideal Gas Law that needs to be recalled is:

$\large {\boxed {PV = nRT} }$

<em>P = Pressure (Pa)</em>

<em>V = Volume (m³)</em>

<em>n = number of moles (moles)</em>

<em>R = Gas Constant (8.314 J/mol K)</em>

<em>T = Absolute Temperature (K)</em>

Let us now tackle the problem !

$\texttt{ }$

<u>Given:</u>

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

<u>Unknown:</u>

Work done on the gas = W = ?

Heat added to the gas = Q = ?

<u>Solution:</u>

<h3>Step A:</h3>

<em>Ideal gas is allowed to expand isothermally:</em>

$P_1V_1 = P_2V_2$

$5.00 \times 2.00 = 3.00 \times V_2$

$V_2 = 10 \div 3$

$V_2 = 3\frac{1}{3} \texttt{ L}$

$\texttt{ }$

<em>Next we will calculate the work done on the gas:</em>

$W_A = -P_2(V_2 - V_1)$

$W_A = -3.00(3\frac{1}{3} - 2.00)$

$W_A = \boxed{-4 \texttt{ L.atm}}$

$\texttt{ }$

<h3>Step B:</h3>

<em>Using the same method as above:</em>

$P_2V_2 = P_3V_3$

$3.00 \times 3\frac{1}{3} = 2.00 \times V_3$

$V_3 = 10 \div 2$

$V_3 = 5 \texttt{ L}$

$\texttt{ }$

<em>Next we will calculate the work done on the gas:</em>

$W_B = -P_3(V_3 - V_2)$

$W_B = -2.00(5 - 3\frac{1}{3})$

$W_B = \boxed{-3\frac{1}{3} \texttt{ L.atm}}$

$\texttt{ }$

<em>Finally we could calculate the total work done and heat added as follows:</em>

$W = W_A + W_B$

$W = -4 + (-3\frac{1}{3})$

$W = -7\frac{1}{3} \texttt{ L.atm}$

$W = -7\frac{1}{3} \times 101.33 \texttt{ J}$

$\boxed{W \approx -743 \textt{ J}}$

$\texttt{ }$

$\Delta U = Q + W$

$0 = Q + (-743)$

$\boxed{Q = 743 \texttt{ J}}$

$\texttt{ }$

<h3>Learn more</h3>

Minimum Coefficient of Static Friction : brainly.com/question/5884009
The Pressure In A Sealed Plastic Container : brainly.com/question/10209135
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Pressure

5 0

4 years ago

A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point ci

MA_775_DIABLO [31]

Answer:

$v_B=3.78\times 10^5\ m/s$

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is $6.63\times 10^{-27}\ kg$

Speed of nucleus at A is $v_A=6.2\times 10^5\ m/s$

Potential at point A, $V_A=1.5\times 10^3\ V$

Potential at point B, $V_B=4\times 10^3\ V$

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

$\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s$

So, the speed at point B is $3.78\times 10^5\ m/s$ .

7 0

3 years ago

A car battery can deliver up to 75 amps of current at a voltage of 12 volts. How much power is this?

Alex_Xolod [135]

A.900 watts That would be your correct answer

4 0

3 years ago

Read 2 more answers