A home electrical system is joined to the electric company's system at the junction of the

There is an electric field near the Earth's surface whose magnitude is about 145 V/m . How much energy is stored per cubic meter

weqwewe [10]

Answer:

u_e = 9.3 * 10^-8 J / m^3 ( 2 sig. fig)

Explanation:

Given:

- Electric Field strength near earth's surface E = 145 V / m

- permittivity of free space (electric constant) e_o = 8.854 *10^-12 s^4 A^2 / m^3 kg

Find:

- How much energy is stored per cubic meter in this field?

Solution:

- The solution requires the energy density stored between earth's surface and the source of electric field strength. The formula for charge density is given by:

u_e = 0.5*e_o * E^2

- Plug in the values given:

u_e = 0.5*8.854 *10^-12 *145^2

u_e = 9.30777 * 10^-8 J/m^3

5 0

3 years ago

Compute the solution to x + 2x + 2x = 0 for Xo = 0 mm, vo = 1 mm/s and write down the closed-form expression for the response.

Nutka1998 [239]

Answer:

β = $\frac{c}{\sqrt{km} }$ = 0.7071 ≈ 1 ( damping condition )

closed-form expression for the response is attached below

Explanation:

Given : x + 2x + 2x = 0 for Xo = 0 mm and Vo = 1 mm/s

computing a solution :

M = 1,

c = 2,

k = 2,

Wn = $\sqrt{\frac{k}{m} }$ = $\sqrt{2}$

next we determine the damping condition using the damping formula

β = $\frac{c}{\sqrt{km} }$ = 0.7071 ≈ 1

from the condition above it can be said that the damping condition indicates underdamping

attached below is the closed form expression for the response

6 0

3 years ago

Q.14

Kay [80]

Answer:

A. optical isolation

Explanation:

well I can't really give a good explanation because I also saw the same question in my exams and option A was the correct answer

6 0

3 years ago

Should aircraft wings have infinite stiffness?

Colt1911 [192]

Answer:

No, they need to be somewhat flexible so that forces such as turbulance don't shear the wing off.

3 0

3 years ago

Read 2 more answers

The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet

alexgriva [62]

Answer:Counter,

0.799,

1.921

Explanation:

Given data

$T_{h_i}=200^{\circ}C$

$T_{h_o}=120^{\circ}C$

$T_{c_i}=100^{\circ}C$

$T_{c_o}=125^{\circ}C$

Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger

Equating Heat exchange

$m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]$

$\frac{m_hc_{ph}}{m_cc_{pc}}$ = $\frac{125-100}{200-120}=\frac{25}{80}=C\left ( capacity rate ratio\right )$

we can see that heat capacity of hot fluid is minimum

Also from energy balance

$Q=UA\Delta T_m=\left ( mc_p\right )_{h}\left ( T_{h_i}-T_{h_o}\right )$

$NTU=\frac{UA}{\left ( mc_p\right )_{h}}$ = $\frac{\left ( T_{h_i}-T_{h_o}\right )}{T_m}$

$T_m=\frac{\left ( 200-125\right )-\left ( 120-100\right )}{\ln \frac{75}{20}}$

$T_m=41.63^{\circ}C$

NTU=1.921

$And\ effectiveness \epsilon =\frac{1-exp\left ( -NTU\left ( 1-c\right )\right )}{1-c\left ( -NTU\left ( 1-c\right )\right )}$

$\epsilon =\frac{1-exp\left ( -1.921\left ( 1-0.3125\right )\right )}{1-0.3125exp\left ( -1.921\left ( 1-0.3125\right )\right )}$

$\epsilon =\frac{1-exp\left ( -1.32068\right )}{1-0.3125exp\left ( -1.32068\right )}$

$\epsilon =\frac{1-0.2669}{1-0.0834}$

$\epsilon =0.799$

5 0

4 years ago