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ioda
3 years ago
5

One watt is equal to which of these values?

Physics
2 answers:
gregori [183]3 years ago
7 0

Answer:

Correct option: 1 joule/second

Explanation:

The Watt is the unit of measure of power of the International System of Units, it is represented by the letter W in uppercase, it owes its name to the Scottish James Watt.

The watt is a unit of power that equals 1 joule per second (1J / s). Or the work of moving a mass of 1 kilogram, a distance of 1 meter in a second of time. Hence the equivalence since force by distance is work and time divided by power. A horsepower equals 775 W. Expressed in units used in electricity, one watt is the electrical power produced by a potential difference of 1 volt and an electric current of 1 ampere (1 voltampere).

dusya [7]3 years ago
3 0
1 W = 1 J/s 
Watt<span> is the unit of measure for power .
</span><span>
</span>Joule is the unit of measure for the energy and the second is the unit of measure for time.


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What acceleration will you give to a 24.5 kg
Ludmilka [50]

Heya!!

For calculate aceleration, lets applicate second law of Newton:

                                                   \boxed{F=ma}

                                                 <u>Δ   Being   Δ</u>

                                             F = Force = 78,3 N

                                            m = Mass = 24,5 kg

                                             a = Aceleration = ?

⇒ Let's replace according the formula and clear "a":

\boxed{a=78,3\ N / 24,5\ kg}

⇒ Resolving

\boxed{a=3.19\ m/s^{2}}

Result:

The aceleration is <u>3,19 meters per second squared (m/s²)</u>

Good Luck!!

4 0
3 years ago
A sphere of radius r = 5cm carries a uniform volume charge density rho = 400 nC/m^3. Q. What is the total charge Q of the sphere
Tanzania [10]

Answer:

The total charge Q of the sphere is 2.094\times10^{-10}\ C.

Explanation:

Given that,

Radius = 5 cm

Charge density J= 400\ nC/m^3

We need to calculate the total charge Q of the sphere

Using formula of charge

q=\rho V

Where, \rho = charge density

V = volume

Put the value into the formula

q=\rho\times(\dfrac{4}{3}\pi r^3)

Put the value into the formula

q=\dfrac{4}{3}\times\pi\times400\times10^{-9}\times(5\times10^{-2})^3

q=2.094\times10^{-10}\ C

Hence, The total charge Q of the sphere is 2.094\times10^{-10}\ C.

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3 0
3 years ago
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The sun continuously radiates energy into space in all directions. Some of the sun's energy is intercepted by the Earth. The ave
8_murik_8 [283]

B. The Earth radiates an amount of energy into space equal to the amount it receives.

Part of the solar energy is reflected by the Earth into space, this is known as albedo. The other part of the energy radiated by the Earth in the form of infrared radiation, is absorbed by the greenhouse gases, which cause most of this infrared radiation to be emitted into space. Therefore, the net flow of energy is zero.

7 0
3 years ago
A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa
bearhunter [10]

Explanation:

Formula which holds true for a leans with radii R_{1} and R_{2} and index refraction n is given as follows.

          \frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

Since, the lens is immersed in liquid with index of refraction n_{1}. Therefore, focal length obeys the following.  

            \frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]  

             \frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

and,       \frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}

or,          f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}

              f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}

                          = 32.4 cm

Using thin lens equation, we will find the focal length as follows.

             \frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}

Hence, image distance can be calculated as follows.

       \frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}

              s_{i} = \frac{fs_{o}}{s_{o} - f}

             s_{i} = \frac{32.4 \times 100}{100 - 32.4}

                       = 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0
3 years ago
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