All categories

2 years ago

Report of invertor to convert 12 volt to 220 volt.

Engineering

1 answer:

Maksim231197 [3]2 years ago

5 0

Sorryyyy I need points

You might be interested in

To ensure that a vehicle crash is inelastic, vehicle safety designers add crumple zones to vehicles. A crumple zone is a part of

spin [16.1K]

Answer:

Explanation:

Answer: With crumple zones at the front and back of most cars, they absorb much of the energy (and force) in a crash by folding in on itself much like an accordion. ... As Newton's second law explains force = Mass x Acceleration this delay reduces the force that drivers and passengers feel in a crash.Sep 30, 2020

5 0

2 years ago

A 1 m wide continuous footing is designed to support an axial column load of 250 kN per meter of wall length. The footing is pla

creativ13 [48]

Answer:

correct option is (A) 0.5

Explanation:

given data

axial column load = 250 kN per meter

footing placed = 0.5 m

cohesion = 25 kPa

internal friction angle = 5°

solution

we know angle of internal friction is 5° that is near to 0°

so it means the soil is almost cohesive soil.

and for a pure cohesive soil

$N_{\gamma }$ = 0

and we know formula for $N_{\gamma }$ is

$N_{\gamma }$ = (Nq - 1 ) × tan(Ф) ..................1

so here Ф is very less $N_{\gamma }$ should be nearest to zero

and its value can be 0.5

so correct option is (A) 0.5

7 0

3 years ago

A 60-cm-high, 40-cm-diameter cylindrical water tank is being transported on a level road. The highest acceleration anticipated i

dlinn [17]

Answer:

$h_{max} = 51.8 cm$

Explanation:

given data:

height of tank = 60cm

diameter of tank =40cm

accelration = 4 m/s2

suppose x- axis - direction of motion

z -axis - vertical direction

$\theta$ = water surface angle with horizontal surface

$a_x =$ accelration in x direction

$a_z =$ accelration in z direction

slope in xz plane is

$tan\theta = \frac{a_x}{g +a_z}$

$tan\theta = \frac{4}{9.81+0}$

$tan\theta =0.4077$

the maximum height of water surface at mid of inclination is

$\Delta h = \frac{d}{2} tan\theta$

$=\frac{0.4}{2}0.4077$

$\Delta h 0.082 cm$

the maximu height of wwater to avoid spilling is

$h_{max} = h_{tank} -\Delta h$

= 60 - 8.2

$h_{max} = 51.8 cm$

the height requird if no spill water is $h_{max} = 51.8 cm$

3 0

3 years ago

1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil

kakasveta [241]

Answer:

a) $n = 1000\,users$ , b) $\Delta t_{min} = \frac{1}{30}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$ , c) $n = 10000000\,users$ , $\Delta t_{min} = \frac{1}{3000}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

Explanation:

a) The total number of users that can be accomodated in the system is:

$n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )$

$n = 1000\,users$

b) The length of the side of each cell is:

$l = \sqrt{1\,km^{2}}$

$l = 1\,km$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{30}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$

c) The total number of users that can be accomodated in the system is:

$n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )$

$n = 10000000\,users$

The length of each side of the cell is:

$l = \sqrt{100\,m^{2}}$

$l = 10\,m$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{3000}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

8 0

3 years ago

3. This material is considered flammable.

almond37 [142]

Answer:

All of the above

Explanation:

Most of materials are flammable because of the chemicals theyre made up of. Most materials that are bought say to keep them at room temature for a reason.

4 0

2 years ago

Read 2 more answers