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spin [16.1K]
2 years ago
13

What is one watt power?​

Physics
2 answers:
QveST [7]2 years ago
8 0

Answer:

1-watt power is defined as the power produced when 1 Joule of work is done for 1 second.

Explanation:

<em>Hope this helps! Please let me know if you need more help or think my answer is incorrect. Brainliest would be MUCH appreciated. Have a wonderful day</em>!

iris [78.8K]2 years ago
4 0

Answer:

<u>In terms of electromagnetism, one watt is the rate at which electrical work is performed when a current of one ampere (A) flows across an electrical potential difference of one volt (V), meaning the watt is equivalent to the volt-ampere (the latter unit, however, is used for a different quantity from the real power of an electrical circuit).</u>

<u>Explanation:</u>

Hope it helps :)

Pls mark brainleist <3

`

`

`

<em>Tori </em>

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A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second; and the m
Damm [24]

Answer

given,                                                

Tension of string is F                                                  

velocity is increased and the radius is not changed.      

the string makes two complete revolutions every second

consider the centrifugal force acting on the stone          

  = \dfrac{mv^2}{r}                          

now centrifugal force is balanced by tension

T =\dfrac{mv^2}{r}                                

From the above expression we can clearly see that tension is directly proportional to velocity and inversely proportional to radius.

When radius is not changing velocity is increasing means tension will also increase in the string.

8 0
3 years ago
Please help me solve this and give an explanation​
Arlecino [84]

Answer:

6.5

Explanation:

Because 1.5+5=6.5

7 0
2 years ago
A closely wound, circular coil with a diameter of 4.30 cm has 470 turns and carries a current of 0.460 A .
Nadusha1986 [10]

Hi there!

a)
Let's use Biot-Savart's law to derive an expression for the magnetic field produced by ONE loop.

dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}

dB = Differential Magnetic field element

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

R = radius of loop (2.15 cm = 0.0215 m)

i = Current in loop (0.460 A)

For a circular coil, the radius vector and the differential length vector are ALWAYS perpendicular. So, for their cross-product, since sin(90) = 1, we can disregard it.

dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{r^2}

Now, let's write the integral, replacing 'dl' with 'ds' for an arc length:
B = \int \frac{\mu_0}{4\pi} \frac{ids}{R^2}

Taking out constants from the integral:
B =\frac{\mu_0 i}{4\pi R^2}  \int ds

Since we are integrating around an entire circle, we are integrating from 0 to 2π.

B =\frac{\mu_0 i}{4\pi R^2}  \int\limits^{2\pi R}_0 \, ds

Evaluate:
B =\frac{\mu_0 i}{4\pi R^2}  (2\pi R- 0) = \frac{\mu_0 i}{2R}

Plugging in our givens to solve for the magnetic field strength of one loop:

B = \frac{(4\pi *10^{-7}) (0.460)}{2(0.0215)} = 1.3443 \mu T

Multiply by the number of loops to find the total magnetic field:
B_T = N B = 0.00631 = \boxed{6.318 mT}

b)

Now, we have an additional component of the magnetic field. Let's use Biot-Savart's Law again:
dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}

In this case, we cannot disregard the cross-product. Using the angle between the differential length and radius vector 'θ' (in the diagram), we can represent the cross-product as cosθ. However, this would make integrating difficult. Using a right triangle, we can use the angle formed at the top 'φ', and represent this as sinφ.  

dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} sin\theta}{r^2}

Using the diagram, if 'z' is the point's height from the center:

r = \sqrt{z^2 + R^2 }\\\\sin\phi = \frac{R}{\sqrt{z^2 + R^2}}

Substituting this into our expression:
dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{(\sqrt{z^2 + R^2})^2} }(\frac{R}{\sqrt{z^2 + R^2}})\\\\dB = \frac{\mu_0}{4\pi} \frac{iRd\vec{l}}{(z^2 + R^2)^\frac{3}{2}} }

Now, the only thing that isn't constant is the differential length (replace with ds). We will integrate along the entire circle again:
B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} \int\limits^{2\pi R}_0, ds

Evaluate:
B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} (2\pi R)\\\\B = \frac{\mu_0 iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}

Multiplying by the number of loops:
B_T= \frac{\mu_0 N iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}

Plug in the given values:
B_T= \frac{(4\pi *10^{-7}) (470) (0.460)(0.0215)^2}{2 ((0.095)^2 + (0.0215)^2)^\frac{3}{2}}} \\\\ =  0.00006795 = \boxed{67.952 \mu T}

5 0
2 years ago
Read 2 more answers
24. While measuring the length of a book, the reading of the scale at one end is 5.0 cm and at the other end is 20.5
il63 [147K]

Answer:

...do

Explanation:

24. While measuring the length of a book, the reading of the scale at one end is 5.0 cm and at the other end is 20.5

cm. What is the length of the book in mm?

25. Explain the modifications

3 0
3 years ago
A machine does 1200 J of work in 1 min. What is the power developed
densk [106]

Answer:

20 watts

Explanation:

Big brain mode activated:

Power=1200J/60sec

Power=20 watts

8 0
3 years ago
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