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denis-greek [22]
3 years ago
11

Which statement provides the best description of a computational model?

Physics
1 answer:
Varvara68 [4.7K]3 years ago
4 0

Answer:

The answer is "Option C".

Explanation:

Computation modeling is used as the software for math, physics, or software engineering in the simulation and study of complicated processes. The software framework incorporates many parameters, that characterize the model under study. It is a computer program, in which the mathematical formula is generated by computers and research complex systems.

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A student connects a small solar panel to a 40 a resistor to make a simple circuit. The solar panel produces a voltage of 2 0.
Nastasia [14]
<h3>Solution for the above question : -</h3>

Ohm's law states that :

  • v = ir

the terms used are :

  • r = resistance
  • v = potential \:  \: difference
  • i  = current

let's solve for electric current :

  • 2 = i \times 40

  • i =  \dfrac{2}{40}

  • i = 0.05 \: A

  • i = 50 \: mA

\mathfrak{good\:  \: luck \:  \: for \:  \: your \:  \: assignment}

8 0
3 years ago
An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o
Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

T=2\pi\sqrt{\dfrac{m}{k}}

m is the mass of object

m=\dfrac{W}{g}

m=\dfrac{2.45}{9.8}

m = 0.25 kg

k=\dfrac{4\pi^2m}{T^2}

k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}

k = 24.09 N/m

or

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

6 0
3 years ago
How can light energy solve our real life problem?
kifflom [539]

Answer:

It gives our light which we need for probably everything.

Explanation:

4 0
3 years ago
Read 2 more answers
A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the
BARSIC [14]

Answer:

Explanation:

This is a simple gravitational force problem using the equation:

F_g=\frac{Gm_1m_2}{r^2} where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.

Filling in:

F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2} I'm going to do the math on the top and then on the bottom and divide at the end.

F_g=\frac{2.4012*10^{40}}{3.721*10^{23}} and now when I divide I will express my answer to the correct number of sig dig's:

Fg= 6.45 × 10¹⁶ N

8 0
3 years ago
Two ice skaters, each with a mass of 72.0 kg, are skating at 5.45 m/s when they collide and stick together. If the angle between
Rudiy27

Answer:

The skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.

Explanation:

To solve the problem it is necessary to go back to the theory of conservation of momentum, specifically in relation to the collision of bodies. In this case both have different addresses, consideration that will be understood later.

By definition it is known that the conservation of the moment is given by:

m_1v_1+m_2v_2=(m_1+m_2)v_f

Our values are given by,

m_1=m_2=72Kg

As the skater 1 run in x direction, there is not component in Y direction. Then,

Skate 1:

v_{x1}=5.45m/s

v_{y1}=0

Skate 2:

v_{x2} = 5.45*cos105= -1.41m/s

v_{y2} = 5.45*sin105 = 5.26m/s

Then, if we applying the formula in X direction:

m_1v_{x1}+m_2v_{x2}=(m_1+m_2)v_{fx}

75*5.45-75*1.41=(75+75)v_{fx}

Re-arrange and solving for v_{fx}

v_{fx}=\frac{4.04}{2}

v_{fx}=2.02m/s

Now applying the formula in Y direction:

m_1v_{y1}+m_2v_{y2}=(m_1+m_2)v_{fy}

0+75*5.25=(75+75)v_{fy}

v_{fy}=\frac{5.25}{2}

v_{fy}=2.63m/s

Therefore the skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.

6 0
3 years ago
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