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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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1 point
WINSTONCH [101]

Answer:

 Mg- 27   means isotope with 12 protons and 15 neutrons.

Also 27 is mass number which express sum of protons and neutrons.

In nucleus one neutrn decays to electron and proton. Mass number remain same but Al-27 nucleus contain 13 protons and 14 neutrons. Electron is ejected out from nucleus.

7 0
3 years ago
A ball is dropped from an aircraft flying at an altitude of 8,848 meters assuming gravity is 9.8m/s what is the total amount of
Dafna11 [192]
In this question, you're determining the time (t) taken for an object to fall from a distance (d).

The equation to represent this is:
Time equals the square root of 2 times the distance divided by the gravitational force of earth.
In equation from it looks like this (there isn't an icon to represent square root so just pretend like there's a square root there):
t = 2d/g (square-rooted)

d = 8,848m and g = 9.8m/s

Now plug in the information we have:
t = 2 x 8,848m/9.8m/s (square-rooted)

The first step is to multiply 2 times 8,848m:
t = 17,696m/9.8m/s (square-rooted)

Now divide 9.8m/s by 17,696m (note that the two m's (meters) cancels out leaving you with only s (seconds):
t = 1805.72s (square-rooted)

Now for the last step, find the square root of the remaining number:
t = 42.5s

So the time it takes the ball to drop from the height (distance) of 8,848 meters, and falling with the gravitational pull of 9.8 meters per second is 42.5 seconds.

I hope this helps :)

7 0
3 years ago
An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get
Blizzard [7]

In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

mV_i = (m-m_O)V_f - m_OV_O

where,

m=Total mass

m_O = Mass of Object

V_i = Velocity before throwing

V_f = Final Velocity

V_O = Velocity of Object

Our values are:

m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg

Solving to find the final speed, after throwing the object we have

V_f=\frac{mV_0+m_TV_O}{m-m_O}

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) 5.3Kg\rightarrow 15m/s

V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}

V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}

V_{f1}=0.8511m/s

B) 7.9Kg\rightarrow 11.2m/s

V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}

V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}

V_{f2} = 2.0173m/s

C) 10.5Kg\rightarrow 7m/s

V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}

V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}

V_{f3} = 3.63478m/s

Therefore the final velocity of astronaut is 3.63m/s

7 0
3 years ago
What are the structures shown below ??
forsale [732]
The body system on the chart
8 0
3 years ago
A flat sheet of ice has a thickness of 2.20 cm. It is on top of a flat sheet of crystalline quartz that has a thickness of 1.50
kolezko [41]

Answer:

Distance_{vaccum}=5.19cm

Explanation:

The speed of light in these mediums shall be lower than that in vacuum thus the total time light needs to cross both the media are calculated as under

Total time = Time taken through ice + Time taken through quartz

Time taken through ice = Thickness of ice / (speed of light in ice)

T_{ice}=\frac{2.20\times 10^{-2} \times \mu _{ice}}{V_{vaccum}}

T_{quartz}=\frac{1.50\times 10^{-2} \times \mu _{quartz}}{V_{vaccum}}

Thus in the same time the it would had covered a distance of

Distance_{vaccum}=Totaltime\times V_{vaccum}\\\\Distance_{vaccum}=10^{-2}[2.20\mu _{ice+1.50\mu _{quartz}}]

we have

\mu _{ice}=1.309\\\\\mu _{quartz}=1.542

Applying values we have

Distance_{vaccum}=10^{-2}[2.20\times 1.309+1.50\times 1.542]

Distance_{vaccum}=5.19cm

6 0
4 years ago
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