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user100 [1]
3 years ago
10

Given: An unbroken length of size 12 AWG conductor is looped inside a junction box so that there is more than 12 inches of free

conductor. The conductor is not spliced and does not terminate on a device. What volume, in cubic inches, must be counted for the looped conductor?
Physics
1 answer:
kakasveta [241]3 years ago
8 0

Answer: 27 cubic inches

Explanation:

Inch capacity for a 12 AWG=2.25

Therefore  

Volume=12*2.25

 =27 cubic inches

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What is a constellation as astronomers define it today? What does it mean when an astronomer says, “I saw a comet in Orion last
MissTica

Explanation:

Constellation: The complete sky has been divided in 88 different areas, in a way we have divided Earth in countries, not necessarily having same shapes and size. These 88 areas are known as constellations. These contains a lot of stars. When we join the brightest stars together we can imagine a shape out of them which is called as Asterism. Most of the people are unaware of this difference. Some of the famous constellations are Orion, Taurus, Gemini, Hydra, Ursa Major etc.

When an astronomer says that there is a comet is in the Orion, he means that a comet is in the boundaries of Orion constellation.

4 0
3 years ago
A skier leaves the horizontal end of a ramp with a velocity of 25.0 m/s and lands 70.0 m from the base of the ramp. How high is
Valentin [98]

Answer:

<em>The end of the ramp is 38.416 m high</em>

Explanation:

<u>Horizontal Motion </u>

When an object is thrown horizontally with an initial speed v and from a height h, it follows a curved path ruled by gravity.

The maximum horizontal distance traveled by the object can be calculated as follows:

\displaystyle d=v\cdot\sqrt{\frac  {2h}{g}}

If the maximum horizontal distance is known, we can solve the above equation for h:

\displaystyle h=\frac  {d^2g}{2v^2}

The skier initiates the horizontal motion at v=25 m/s and lands at a distance d=70 m from the base of the ramp. The height is now calculated:

\displaystyle h=\frac  {70^2\cdot 9.8}{2\cdot 25^2}

\displaystyle h=\frac  {4900\cdot 9.8}{2\cdot 625}

h= 38.416 m

The end of the ramp is 38.416 m high

8 0
3 years ago
A student pulls horizontally on a 12 kg box, which then moves horizontally with an acceleration of 0.2 m/s^2. If the student use
polet [3.4K]
The net force of the object is equal to the force applied minus the force of friction. 
                         Fnet = ma = F - Ff
                           12 kg x 0.2 m/s² = 15 N - Ff
The value of Ff is 12.6 N. This force is equal to the product of the normal force which is equal to the weight in horizontal surface and the coefficient of friction.
                             Ff = 12.6 N = k(12 kg)(9.81 m/s²)
The value of k is equal to 0.107. 
7 0
3 years ago
A 50-kg block is at rest on a 15o slope. A force of 250 N is acting on the block up the slope parallel to it. If the block does
Iteru [2.4K]

The minimum value of the coefficient of static friction between the block and the slope is 0.53.

<h3>Minimum coefficient of static friction</h3>

Apply Newton's second law of motion;

F - μFs = 0

μFs = F

where;

  • μ is coefficient of static friction
  • Fs is frictional force
  • F is applied force

μ = F/Fs

μ = F/(mgcosθ)

μ = (250)/(50 x 9.8 x cos15)

μ = 0.53

Thus, the minimum value of the coefficient of static friction between the block and the slope is 0.53.

Learn more about coefficient of friction here: brainly.com/question/20241845

#SPJ1

6 0
2 years ago
In an electric field, 0.90 joule of work is required to bring 0.45 coulomb of charge from point a to point
jarptica [38.1K]
The difference in electric potential energy between the two points is
\Delta U = q \Delta V
where q is the magnitude of the charge and \Delta V is the electric potential difference.

But for energy conservation, the difference in electric potential energy \Delta U between the two points is equal to the work done to move the charge between A and B:
W=\Delta U
so we have
W=q \Delta V

and by substituting the numbers of the problem, we find the value of \Delta V:
\Delta V =  \frac{W}{q}= \frac{0.90 J}{0.45 C}=2 V
3 0
3 years ago
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