Answer:
140.83 W
Explanation:
From the question given above, the following data were obtained:
Height (h) = 26 m
Time (t) = 12 s
Mass (m) = 6.50 Kg
Power (P) =?
Power is simply defined as the rate at which energy is used up. Mathematically, it is expressed as:
Power (P) = Energy (E) / Time (t)
But:
Energy (E) = mass (m) × acceleration due to gravity (g) × height (h)
E = mgh
P = E/t
P = mgh / t
With the above formula i.e
P = mgh / t
We can obtain the power supplied by Jill's muscles as follow:
Height (h) = 26 m
Time (t) = 12 s
Mass (m) = 6.50 Kg
Acceleration due to gravity (g) = 10 m/s²
Power (P) =?
P = mgh / t
P = 6.5 × 10 × 26 / 12
P = 140.83 W
Therefore, the power supplied by Jill's muscles is 140.83 W
Answer:
T=1.384×10⁶seconds
Explanation:
Given data
p (Intensity)=1.30 kw/m²
E (Energy)=1.8×10⁹ J
A (Area)=1.00 m²
T (Time required)=?
Solution
E=PT ................eq(i)
where E is energy
P is radiation power
T is time
Radiating Power is given as
P=pA
Where p is intensity
A is Area
Put P=pA in eq(i) we get
E=pAT
T=E/pA
The variable x would be 4
Answer:
Option (3)
Explanation:
Nicolaus Copernicus was an astronomer from Poland, who was born on the 19th of February in the year 1473. He played a great role in the field of modern astronomy.
He was the person who contributed to the heliocentric theory. This theory describes the position of the sun in the middle of the universe, and all the planets move around the sun. This theory was initially not accepted, and after about a century it was widely accepted.
This theory describes the present-day motion of the planets around the sun in the solar system. This theory replaced the geocentric theory.
Thus, the correct answer is option (3).
To solve this problem we will use the work theorem, for which we have that the Force applied on the object multiplied by the distance traveled by it, is equivalent to the total work. From the measurements obtained we have that the width and the top are 14ft and 7ft respectively. In turn, the bottom of the tank is 15ft. Although the weight of the liquid is not given we will assume this value of (Whose variable will remain modifiable until the end of the equations subsequently presented to facilitate the change of this, in case be different). Now the general expression for the integral of work would be given as
Basically under this expression we are making it difficult for the weight of the liquid multiplied by the area (Top and widht) under the integral of the liquid path to be equivalent to the total work done, then replacing
Therefore the total work in the system is