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2 years ago

The centre of the circumstancribing circle of a triangle can be found by using the

Engineering

1 answer:

liraira [26]2 years ago

7 0

Answer:

perpendicular bisectors of each side of the triangle I believe

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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

Bridging are members installed periodically between joists to ensure which of the following?

Pie

Answer:

Distributes a floor load or weight

Explanation:

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3 years ago

Omg help mr idk what to say ahhh

kap26 [50]

Explanation:

ответ на фото !!!!!!

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3 years ago

Read 2 more answers

An ideal vapor-compression refrigeration cycle operates at steady state with Refrigerant 134a as the working fluid. Saturated va

Ksju [112]

Explanation:

Note: Refer the diagram below

Obtaining data from property tables

State 1:

$\left.\begin{array}{l}P_{1}=1.25 \text { bar } \\\text { Sat - vapour }\end{array}\right\} \begin{array}{l}h_{1}=234.45 \mathrm{kJ} / \mathrm{kg} \\S_{1}=0.9346 \mathrm{kJ} / \mathrm{kgk}\end{array}$

State 2:

$\left.\begin{array}{l}P_{2}=5 \text { bor } \\S_{2}=S_{1}\end{array}\right\} \quad h_{2}=262.78 \mathrm{kJ} / \mathrm{kg}$

State 3:

$\left.\begin{array}{l}P_{3}=5 \text { bar } \\\text { Sat }-4 q\end{array}\right\} h_{3}=71-33 \mathrm{kJ} / \mathrm{kg}$

State 4:

Throttling process $h_{4}=h_{3}=71.33 \mathrm{kJ} / \mathrm{kg}$

(a)

Magnitude of compressor power input

$\dot{w}_{c}=\dot{m}\left(h_{2}-h_{1}\right)=\left(8 \cdot 5 \frac{\mathrm{kg}}{\min } \times \frac{1 \mathrm{min}}{\csc }\right)(262.78-234 \cdot 45)\frac{kj}{kg}$