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3 years ago

8

Solve the compound inequality. 3x − 4 > 5 or 1 − 2x ≥ 7

1 answer:

Kaylis [27]3 years ago

3 0

X ≤ −3 or x > 3
inequality form

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Cómo se llama el componente, que permite abrir o cerrar un circuito eléctrico

Korolek [52]

Creo haber leído sobre esto..Un “interruptor” yo diría

5 0

2 years ago

An electrical current of 700 A flows through a stainlesssteel cable having a diameter of 5 mm and an electricalresistance of 610

KatRina [158]

Answer:

778.4°C

Explanation:

I = 700

R = 6x10⁻⁴

we first calculate the rate of heat that is being transferred by the current

q = I²R

q = 700²(6x10⁻⁴)

= 490000x0.0006

= 294 W/M

we calculate the surface temperature

Ts = T∞ + $\frac{q}{h\pi Di}$

Ts = $30+\frac{294}{25*\frac{22}{7}*\frac{5}{1000} }$

$Ts=30+\frac{294}{0.3928} \\$

$Ts =30+748.4\\Ts = 778.4$

The surface temperature is therefore 778.4°C if the cable is bare

6 0

2 years ago

A piston-cylinder apparatus has a piston of mass 2kg and diameterof

iragen [17]

Answer:

M =2.33 kg

Explanation:

given data:

mass of piston - 2kg

diameter of piston is 10 cm

height of water 30 cm

atmospheric pressure 101 kPa

water temperature = 50°C

Density of water at 50 degree celcius is 988kg/m^3

volume of cylinder is $V = A \times h$

$= \pi r^2 \times h$

$= \pi 0.05^2\times 0.3$

mass of available in the given container is

$M = V\times d$

$= volume \times density$

$= \pi 0.05^2\times 0.3 \times 988$

M =2.33 kg

6 0

3 years ago

A 11.5 nC charge is at x = 0cm and a -1.2 nC charge is at x = 3 cm ..At what position or positions on the x-axis is the electric

diamong [38]

Answer:

Explanation:

Given

$q_1=11.5\ nC$ charge is placed at $x=0\ cm$

another charge of $q_2=-1.2\ nC$ is at $x=3\ cm$

We know that Electric field due to positive charge is away from it and Electric field due to negative charge is towards it.

so net electric field is zero somewhere beyond negatively charged particle

Electric Field due to $q_2$ at some distance r from it

$E_2=\frac{kq_2}{r^2}$

Now Electric Field due to $q_1$ is

$E_1=\frac{kq_1}{(3+r)^2}$

Now $E_1+E_2=0$

$\frac{k\times 11.5}{(r+3)^2}\frac{k\times (-1.2)}{r^2}=0$

$\frac{3+r}{r}=(\frac{11.5}{1.2})^{0.5}$

$\frac{3+r}{r}=3.095$

thus $r=1.43\ cm$

Thus Electric field is zero at some distance r=1.43 cm right of $q_2$

3 0

2 years ago

Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially t

zepelin [54]

Answer:

attached below

Explanation:

4 0

3 years ago