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MaRussiya [10]
3 years ago
7

One watt is A. equal to one kilogram per second B. the power that can be delivered by an average horse C. the same as one-foot p

ound per second D. the proper replacement unit for one joule per second E. the same as one horsepower per second
Physics
1 answer:
Naddik [55]3 years ago
8 0

Answer:

D. the proper replacement unit for one joule per second

Explanation:

When energy is divided by the time the energy was used we get power

P=\dfrac{E}{t}\\\Rightarrow P=\dfrac{mv^2}{t}\\\Rightarrow P=\dfrac{kg(\dfrac{m^2}{s^2})}{s}\\\Rightarrow P=kg(\dfrac{m^2}{s^2})}\times \dfrac{1}{s}\\\Rightarrow P=\dfrac{kgm^2}{s^3}

kg\dfrac{m^2}{s^2}=Joule

P=\dfrac{kgm^2}{s^3}

So, the answer is D. the proper replacement unit for one joule per second

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Answer:

because a raduis is half of 25% of a cicrle.

Explanation:

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3 years ago
g Design an experiment you can use to determine the mass of the metal cylinder. When you explain your experiment, be sure to men
kumpel [21]

Answer:

m = \frac{k}{g} x,

graph of x vs m

Explanation:

For this exercise, the simplest way to determine the mass of the cylinder is to take a spring and hang the mass, measure how much the spring has stretched and calculate the mass, using the translational equilibrium equation

              F_e -W = 0

              k x = m g

              m = \frac{k}{g} x

We are assuming that you know the constant k of the spring, if it is not known you must carry out a previous step, calibrate the spring, for this a series of known masses are taken and hung by measuring the elongation (x) from the equilibrium position, with these data a graph of x vs m is made to serve as a spring calibration.

  In the latter case, the elongation measured with the cylinder is found on the graph and the corresponding ordinate is the mass

3 0
3 years ago
If an automobile's velocity changes from 25 m/s to 15m/s in 2s, then what is its acceleration?
tekilochka [14]
Hey!

u=25m/s
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4 years ago
Challenge! A marshmallow is dropped from a 5-meter high pedestrian bridge and 0.83 seconds later, it lands right on the head of
WINSTONCH [101]

a₀).  You know ...
         -- the object is dropped from 5 meters
             above the pavement;
         --  it falls for 0.83 second.

a₁).  Without being told, you assume ...
         -- there is no air anyplace where the marshmallow travels,
             so it free-falls, with no air resistance;
         -- the event is happening on Earth,
            where the acceleration of gravity is  9.81 m/s² .

b).  You need to find how much LESS than 5 meters
       the marshmallow falls in 0.83 second.
    
c).  You can use whatever equations you like.
       I'm going to use the equation for the distance an object falls in
       ' T ' seconds, in a place where the acceleration of gravity is ' G '.

d).  To see how this all goes together for the solution, keep reading:


The distance that an object falls in ' T ' seconds
when it's dropped from rest is

                                 (1/2 G) x (T²) .

On Earth, ' G ' is roughly  9.81 m/s², so in 0.83 seconds,
such an object would fall

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It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was

                         (5.00 - 3.38)  =  1.62 meters

above the pavement.  That's where the head of the unsuspecting
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podryga [215]
<span>you must  first select an axis of rotation about which to calculate moment arms and torques. </span>
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