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
Answer:
V = 20.5 m/s
Explanation:
Given,
The mass of the cart, m = 6 Kg
The initial speed of the cart, u = 4 m/s
The acceleration of the cart, a = 0.5 m/s²
The time interval of the cart, t = 30 s
The final velocity of the cart is given by the first equation of motion
v = u + at
= 4 + (0.5 x 30)
= 19 m/s
Hence the final velocity of cart at 30 seconds is, v = 19 m/s
The speed of the cart at the end of 3 seconds
V = 19 + (0.5 x 3)
= 20.5 m/s
Hence, the final velocity of the cart at the end of this 3.0 second interval is, V = 20.5 m/s
Answer:
λ = 3.2 x 10⁻⁷ m = 320 nm
Explanation:
The relationship between the velocity of electromagnetic waves (UV rays) and the their frequency is:
v = fλ
where,
v = c = speed of the electromagnetic waves (UV rays) = speed of light
c = 3 x 10⁸ m/s
f = frequency of the electromagnetic waves (UV rays) = 9.38 x 10¹⁴ Hz
λ = wavelength of the electromagnetic waves (UV rays) = ?
Therefore, substituting the values in the relation, we get:
3 x 10⁸ m/s = (9.38 x 10¹⁴ Hz)(λ)
λ = (3 x 10⁸ m/s)/(9.38 x 10¹⁴ Hz)
<u>λ = 3.2 x 10⁻⁷ m = 320 nm</u>
So, the radiation of <u>320 nm</u> wavelength is absorbed by Ozone.
Answer:

Explanation:
The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.
Also, we know that the centripetal force of an object describing a circular motion is given by:

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.
Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So
and
(Since
). Then, we get:

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).