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4 years ago

An infinite line of charge with charge density λ1 = 0.6 μC/cm is aligned with the y-axis.

Physics

1 answer:

Ganezh [65]4 years ago

8 0

Answer:

1440 × 10⁴ N/C

Explanation:

Data provided in the question:

Charge density λ1 = 0.6 μC/cm = 6 × 10⁻⁵ C/m

a = 7.5 cm = 0.075 m

Now,

Electric field due to a line charge at a distance 'a' is given as:

Ex(P) = $\frac{\lambda}{2\pi\epsilon_0a}$

also,

we know

$\frac{1}{4\pi\epsilon_0}$ = 9 × 10⁹ Nm²/C²

Thus,

we have

Ex(P) = $\frac{2}{2}\times\frac{\lambda}{2\pi\epsilon_0a}=\frac{2\lambda}{4\pi\epsilon_0a}$

therefore,

Ex(P) = $\frac{2\times6\times10^{-5}\times9\times10^9}{0.075}$

= 1440 × 10⁴ N/C

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An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of the compression process,

Sedbober [7]

Answer:

$a.T_3=1723.8kPa\\b.n=0.563\\c.MEP=674.95kPa$

Explanation:

a. Internal energy and the relative specific volume at $s_1$ are determined from A-17: $u_1=214.07kJ/kg, \ \alpha_r_1=621.2$ .

The relative specific volume at $s_2$ is calculated from the compression ratio:

$\alpha_r_2=\frac{\alpha_r_1}{r}\\=\frac{621.2}{16}\\=38.825$

#from this, the temperature and enthalpy at state 2, $s_2$ can be determined using interpolations $T_2=862K$ and $h_2=890.9kJ/kg$ . The specific volume at $s_1$ can then be determined as:

$\alpha_1=\frac{RT_1}{P_1}\\\\=\frac{0.287\times 300}{95} m^3/kg\\0.906316m^3/kg$

Specific volume, $s_2$ :

$\alpha_2=\frac{\alpha_1}{r}\\=\frac{0.906316}{16}m^3/kg\\=0.05664m^3/kg$

The pressures at $s_2 \ and\ s_3$ is:

$P_2=P_3=\frac{RT_2}{\alpha_2}\\\\=\frac{0.287\times862}{0.05664}\\=4367.06kPa$

.The thermal efficiency=> maximum temperature at $s_3$ can be obtained from the expansion work at constant pressure during $s_2-s_3$

$\bigtriangleup \omega_2_-_3=P(\alpha_3-\alpha_2)\\R(T_3-T_2)=P\alpha(r_c-1)\\T_3=T_2+\frac{P\alpha_2}{R}(r_c-1)\\\\=(862+\frac{4367\times 0.05664}{0.287}(2-1))K\\=1723.84K$

b.Relative SV and enthalpy at $s_3$ are obtained for the given temperature with interpolation with data from A-17 : $a_r_3=4.553 \ and\ h_3=1909.62kJ/kg$

Relative SV at $s_4$ is

$a_r_4=\frac{r}{r_c}\alpha _r_3$

= $=\frac{16}{2}\times4.533\\=36.424$

Thermal efficiency occurs when the heat loss is equal to the internal energy decrease and heat gain equal to enthalpy increase;

$n=1-\frac{q_o}{q_i}\\=1-\frac{u_4-u_1}{h_3-h_2}\\=1-\frac{65903-214.07}{1909.62-890.9}\\=0.563$

Hence, the thermal efficiency is 0.563

c. The mean relative pressure is calculated from its standard definition:

$MEP=\frac{\omega}{\alpa_1-\alpa_2}\\=\frac{q_i-q_o}{\alpha_1(1-1/r)}\\=\frac{1909.62-890.9-(65903-214.7)}{0.90632(1-1/16)}\\=674.95kPa$

Hence, the mean effective relative pressure is 674.95kPa

3 0

3 years ago

A uniformly charged sphere has a total charge of 300uc and a radius of 8cm. Find the electric field density at A point 16cm from

s2008m [1.1K]

E = kQ 

(r + h)²

where,

k = 9 × 10^9Nm²C^-2

Q = total charge, 300uC = 300 × 10^ -6C

r = 8 × 10^ -2m

h = 16 × 10^ -2m

then,

E = 9e9 * 300e^-6 

(8e^-2 + 16e^-2)²

E = 4687500N/C

6 0

2 years ago

A horizontal 810-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 55 N applied tangentia

Sloan [31]

Answer:

576 joules

Explanation:

From the question we are given the following:

weight = 810 N

radius (r) = 1.6 m

horizontal force (F) = 55 N

time (t) = 4 s

acceleration due to gravity (g) = 9.8 m/s^{2}

K.E = 0.5 x MI x ω^{2}

where MI is the moment of inertia and ω is the angular velocity

MI = 0.5 x m x r^2

mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg

MI = 0.5 x 82.65 x 1.6^{2}

MI = 105.8 kg.m^{2}

angular velocity (ω) = a x t

angular acceleration (a) = torque ÷ MI

where torque = F x r = 55 x 1.6 = 88 N.m

a= 88 ÷ 105.8 = 0.83 rad /s^{2}

therefore

angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s

K.E = 0.5 x MI x ω^{2}

K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules

6 0

3 years ago

Please help !! In the diagram, q1 = +0.00200 C, q2 = 0.00180 C, and q3 = +0.00830 C. the net force on q2 is zero. how far is q2

VikaD [51]

Answer:

2.03715

Explanation:

$32364=8.99\cdot 10^9\cdot \frac{0.00180\cdot 0.00830}{2.03715^2}$

4 0

3 years ago

A common misconception is that an object always moves when a force acts on it. Why is this statement incorrect? Explain the conc

dsp73

Answer:

The statement is incorrect because, a force acting on an object does not necessarily have to produce motion.

People have the misconception that when a force acts on an object it always produces motion

Explanation:

The statement is incorrect because, a force acting on an object does not necessarily have to produce motion. It could be in static equilibrium where the net force is zero and produces not motion. The body could also be in dynamic equilibrium when no net force acts on it moving at a constant velocity. But here we are concerned with static equilibrium since the body does not move at all.

People have the misconception that when a force acts on an object it always produces motion and, we have seen from the above tat its not always true.

3 0

3 years ago