Computer simulation is useful because it helps in the prediction of what will likely happen in the future using data from past events.
<h3>What is computer simulation?</h3>
- This is the use of computer models to represents a hypothetical scenarios that are likely to be obtained in the real world.
Computer simulations are useful in studying phenomena in the universe because they help us to achieve the followings;
- It helps in the prediction of what will likely happen in the future using data from past events.
- It saves cost and time of carrying out actual experiments.
- It can help prevent a disaster that may occur in the future.
Learn more about computer simulations here: brainly.com/question/22214039
By Newton's second law, the net force on the object is
∑ <em>F</em> = <em>m</em> <em>a</em>
∑ <em>F</em> = (2.00 kg) (8 <em>i</em> + 6 <em>j</em> ) m/s^2 = (16.0 <em>i</em> + 12.0 <em>j</em> ) N
Let <em>f</em> be the unknown force. Then
∑ <em>F</em> = (30.0 <em>i</em> + 16 <em>j</em> ) N + (-12.0 <em>i</em> + 8.0 <em>j</em> ) N + <em>f</em>
=> <em>f</em> = (-2.0 <em>i</em> - 12.0 <em>j</em> ) N
Answer:
Explanation:
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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: Although low frequency sound travels further than high-frequency sound, calls at higher frequencies give the bats more detailed information--such as size, range, position, speed and direction of a prey's flight. Thus, these sounds are used more often.
Explanation:
The distance an object falls from rest through gravity is
D = (1/2) (g) (t²)
Distance = (1/2 acceleration of gravity) x (square of the falling time)
We want to see how the time will be affected
if ' D ' doesn't change but ' g ' does.
So I'm going to start by rearranging the equation
to solve for ' t '. D = (1/2) (g) (t²)
Multiply each side by 2 : 2 D = g t²
Divide each side by ' g ' : 2 D/g = t²
Square root each side: t = √ (2D/g)
Looking at the equation now, we can see what happens to ' t ' when only ' g ' changes:
-- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'
and smaller 'g' ==> longer 't' .--
They don't change by the same factor, because 1/g is inside the square root. So 't' changes the same amount as √1/g does.
Gravity on the surface of the moon is roughly 1/6 the value of gravity on the surface of the Earth.
So we expect ' t ' to increase by √6 = 2.45 times.
It would take the same bottle (2.45 x 4.95) = 12.12 seconds to roll off the same window sill and fall 120 meters down to the surface of the Moon.