The power of man performing 500 J of work in 8 seconds is 62.5 J/s.
Power can be defined as the pace at which work is completed in a given amount of time.
Horsepower is sometimes used to describe the power of motor vehicles and other machinery.
The pace at which work is done on an item is defined as its power. Power is a temporal quantity.
Which is connected to how quickly a project is completed.
The power formula is shown below.
Power = Energy / Time
Power = E / T
Because the standard metric unit for labour is the Joule and the standard metric unit for time is the second, the standard metric unit for power is a Joule / second, defined as a Watt and abbreviated W.
Here we have given Energy as 500 J and Time as 8 second.
Power = Energy / Time
Power = 500 / 8 Joule / sec
Power = 250 / 4 Joule / sec
Power = 125 / 2 Joule / sec
Power = 62.5 Joule / sec or 62.5 watt
Power came out to be 62.5 J/s when the man performed 500 Joule of work in 8 seconds.
So we can conclude that the power in the Energy transmitted per unit of time, and can be find out by dividing Energy by time. In our case the Power came out to be 62.5 Joule / Second.
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Answer:
The answer is a=b=c=3.833 cm
Explanation:
Lets call the variables a=side a b=side b c=side c
We have that the formula of the equilateral triangle for its height is:
1)h=(1/2)*root(3)*a
2) If we resolve the equation we have
2.1)2h=root(3)*a
2.2)(2h/root(3))=a
3) After the replacement of each value we have that
a=2*3.32/1.73205
a=3.833 cm
And we know that the equilateral triangle has the same value for each side so a=b=c=3.833 cm
Mass is <span>is a dimensionless quantity representing the amount of matter in a particle or object. The more mass something has, the more energy is used to lift it.</span>
Answer:
B. chemical potential energy
Explanation:
If an object, such as a ball is lifted above the ground it has gravitational potential energy.
Answer:
The electric potential at the center of the meter stick is 54 KV.
Explanation:
Electric potential (V) is given as:
i.e V = 
Where: k is the Coulomb constant, q is the charge and r is the distance.
Given: q = 3.0 μC = 3.0 x 
C, r = 0.5 m
So that,
V = 
= 
V = 54000
= 54 000 volts
The electric potential at the center of the meter stick is 54 KV.
