F=ma
F = 148×(85-35)÷20
F = 148×(50÷20)
F = 148×2.5
F = 370N
Answer:
1985kg
Explanation:
assuming that
pi =3.14
oil density = 950kg/ cubic meter
g= 9.8m/s

Answer:
Explanation:
Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.
To objective is to find the:
(i) required heat exchanger area.
(ii) flow rate to be maintained in the evaporator.
Given that:
water temperature = 300 K
At a reasonable depth, the water is cold and its temperature = 280 K
The power output W = 2 MW
Efficiency
= 3%
where;



However, from the evaporator, the heat transfer Q can be determined by using the formula:
Q = UA(L MTD)
where;

Also;




LMTD = 4.97
Thus, the required heat exchanger area A is calculated by using the formula:

where;
U = overall heat coefficient given as 1200 W/m².K

The mass flow rate:

Answer:
Magnets come in a variety of shapes and one of the more common is the horseshoe (U) magnet. The horseshoe magnet has north and south poles just like a bar magnet but the magnet is curved so the poles lie in the same plane. The magnetic lines of force flow from pole to pole just like in the bar magnet.
Answer:
Explanation:
Let's look at a mathematical representation of this. The equation for tis is just a souped up version of Newton's 2nd Law:
F - f = ma. It an object is moving at a constant speed, the acceleration of that object is 0. That changes this equation to
F = f which states that the applied Force equals the frictional force, choice a.