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dusya [7]
3 years ago
9

A(n) ? is a premises wiring system whose power is derived from a source of electric energy or equipment other than a service. Su

ch systems have no direct connection from circuit conductors of one system to circuit conductors of another system, other than those established through bonding or grounding connections.
Physics
1 answer:
Evgen [1.6K]3 years ago
3 0

Answer: Separately derived system

Explanation: A separately derived system is used to describe a premise wiring system whose power is derived from a source of electrical energy such as transformer, solar photovoltaic cell or generator. A separately derived system has no direct connection to any conductor from another system or doesn't generate it's power from any direct connection to a conductor from another system or source except those from established from bonding or grounding connections. Separately derived systems usually generate it's power on it's own.

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Based on the principles of convection, conduction and thermal radiation, which scenario below is most similar to the following s
Ksivusya [100]

The situation (heat going through the ceiling) describes
conduction ...
heat going from one place to another by
soaking through some material.

A).  This is the one.  Heat goes from from the marshmallow
to your hand by soaking through the wire.   This is conduction too.

B).  No.  The heat in the room goes from the floor to the ceiling
because the warm air rises and carries it there.  This is convection.

C).  No.  There's nothing for the heat to soak through between
the sun and the roof, and nothing that can move from the sun
to the roof and bring the heat with it.  This is radiation.

D).  No.  Cold water sinks from the surface to the bottom because
warm water rose from the bottom to the surface, taking heat with it. 
This is convection.

5 0
3 years ago
Read 2 more answers
True or false question please help me out
Len [333]

Answer:

thats true

Explanation:

7 0
2 years ago
In 1865, Jules Verne proposed sending men to the Moon by firing a space capsule from a 220-m-long cannon with final speed of 10.
Sidana [21]

Answer:

The unrealistically large acceleration experienced by the space travelers during their launch is 2.7 x 10⁵ m/s².

How many times stronger than gravity is this force? 2.79 x 10⁴ g.

Explanation:

given information:

s = 220 m

final speed, vf = 10.97 km/s = 10970 m/s

g = 9.8 m/s²

he unrealistically large acceleration experienced by the space travelers during their launch

vf² = v₀²+2as, v₀ = 0

vf² = 2as

a =vf²/2s

  = (10970)²/(2x220)

  = 2.7 x 10⁵ m/s²

Compare your answer with the free-fall acceleration

a/g = 2.7 x 10⁵/9.8

a/g = 2.79 x 10⁴

a = 2.79 x 10⁴ g

7 0
3 years ago
You set your stationary bike on a high 80-N friction-like resistive force and cycle for 30 min at a speed of 8.0 m/s . Your body
stellarik [79]

A) The change in internal chemical energy is 1.15\cdot 10^7 J

B) The time needed is 1 minute

Explanation:

First of all, we start by calculating the power output of you and the bike, given by:

P=Fv

where

F = 80 N is the force that must be applied in order to overcome friction and travel at constant speed

v = 8.0 m/s is the velocity

Substituting,

P=(80)(8.0)=640 W

The energy output is related to the power by the equation

P=\frac{E}{t}

where:

P = 640 W is the power output

E is the energy output

t = 30 min \cdot 60 = 1800 s is the time elapsed

Solving for E,

E=Pt=(640)(1800)=1.15\cdot 10^6 J

Since the body is 10% efficient at converting chemical energy into mechanical work (which is the output energy), this means that the change in internal chemical energy is given by

\Delta E = \frac{E}{0.10}=\frac{1.15\cdot 10^6}{0.10}=1.15\cdot 10^7 J

B)

From the previous part, we found that in a time of

t = 30 min

the amount of internal chemical energy converted is

E=1.15\cdot 10^7 J

Here we want to find the time t' needed to convert an amount of chemical energy of

E'=3.8\cdot 10^5 J

So we can setup the following proportion:

\frac{t}{E}=\frac{t'}{E'}

And solving for t',

t'=\frac{E't}{E}=\frac{(3.8\cdot 10^5)(30)}{1.15\cdot 10^7}=1 min

Learn more about power and energy:

brainly.com/question/7956557

#LearnwithBrainly

3 0
2 years ago
Calculate the orbital speed (in m/s) of a satellite that circles the Earth with a time period of 12.00 hours. The mass of the Ea
Hoochie [10]

Answer:

v = 3869 m/s

Explanation:

As we know that the orbital speed of the satellite is given as

v = \sqrt{\frac{GM}{r}}

also we know that

time period of the revolution is given as

T = \frac{2\pi r}{v}

now from above equation we know that

T = \frac{2\pi (\frac{GM}{v^2})}{v}

T = \frac{2\pi GM}{v^3}

so we will have

v = (\frac{2\pi GM}{T})^{1/3}

now plug in all data in this equation

v = (\frac{2\pi (6.67 \times 10^{-11})(5.97 \times 10^{24})}{12 \times 3600})^{1/3}

v = 3869 m/s

3 0
2 years ago
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