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Katen [24]
2 years ago
10

Why are computer simulations useful in studying phenomena in the universe?

Physics
1 answer:
kenny6666 [7]2 years ago
5 0

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

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How long will it take for a car at rest to accelerate at 7m/s^2 to a speed of 45 m/s
Gennadij [26K]
Speed/acceleration = time
45/7
6.4
You have to round the decimal to 6.4
8 0
3 years ago
Tala wants to create a sound echo. She can create the echo using a pane of glass or a carpet square. Which one does she choose?
Alisiya [41]
Carpet asorbs sound, so glass should make it echo.
7 0
3 years ago
Which FBD would represent a car moving right with a motor force of 250 N, and force of friction of 750N, a weight of 8500N and a
babymother [125]

Answer:

Option C

Explanation:

Given that

Motor force is 250 N

Force of friction is 750 N

Weight is 8500 N

And, the normal force is 8500 N

Now based on the above information

Here length of the rector shows the relative magnitude forward force i.e. 250 N i..e lower than the frictional force i.e. backward and weight i.e. 8500 would be equivalent to the normal force

8 0
3 years ago
A block of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7
maxonik [38]
First, let's find the speed v_i of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
p_i = m_1 v_1
After the collision, the two blocks stick together and so now they have mass m_1 +m_2 and they are moving with speed v_i:
p_f = (m_1 + m_2)v_i
For conservation of momentum
p_i=p_f
So we can write
m_1 v_1 = (m_1 +m_2)v_i
From which we find
v_i =  \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force \mu (m_1+m_2)g. The work done by the frictional force to stop the two blocks is
\mu (m_1+m_2)g  d
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
\frac{1}{2} (m_1+m_2)v_i^2
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g  d
From which we can find the value of the coefficient of kinetic friction:
\mu =  \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49
3 0
3 years ago
What is happening in the brain, because people are so rude?
Masteriza [31]
That's a good question.
8 0
3 years ago
Read 2 more answers
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