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lys-0071 [83]
2 years ago
6

What is one advantage of using primary sources when doing research on an

Physics
1 answer:
WITCHER [35]2 years ago
5 0

One advantage of using primary sources when doing research on an experiment is, it contain more details about the experiment. Thus, option D is right .

To find the answer, we have to know more about the Primary sources.

<h3>What is the advantages of primary sources?</h3>
  • Primary sources are the direct or first-hand excerpts that scientists have independently written based on their experiments in order to support their research.
  • Utilizing original sources encourages analytical and critical thought in relation to the research.
  • It aids in exploring from various angles, which leads to the discovery of extra facts.
  • Primary data aids in navigating the conflicts. Since it is a direct resource related to the experiment, it serves as evidence for the data.

Thus, we can conclude that, Primary sources contain more details about the experiment.

Learn more about the Primary sources here:

brainly.com/question/24190082

#SPJ9

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PLZZZ HELP!!! What is the main difference between a permanent magnet and a temporary magnet?
HACTEHA [7]
The domains in a temporary magnet easily lose alignment, but the domains in a permanent magnet keep their alignment.
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A roulette wheel with a 1.0m radius reaches a maximum angular speed of 18 rad/s before it stops 35 revolutions ( 220 rad ) after
Alex17521 [72]
Max ang. speed(u) = 18 rad/s
final ang. speed(v) = 0
ang. displacement(s) = 220 rad

ang. acceleration = (v^2 - u^2)/2s = -18^2 / 2*220 = -0.7364 rad/s^2

v = u +at
0 = 18 - 0.7364t
t = 18/0.7364
t = 24.44 seconds
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3 years ago
If a car increases its velocity from zero to 60 m/s in 10 seconds, its acceleration is A. 600 m/s2 B. 60 m/s2 C. 3 m/s2 D. 6 m/s
enot [183]

Answer:

a

Explanation:

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4 years ago
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Un alumno menciona que al abrir la ventana de su casa sintió cómo el frío ingresaba a su cuerpo. Menciona cuál es la verdadera r
stepan [7]

Answer:

My believe the answer is

A.) or B.)

Explanation:

Here is why I think A is the answer.

If we use the process of elimination, it would look like this,

a) Porque el aire tiene una temperatura menor que la de su cuerpo; por eso se propaga más rápido.

<em>This makes sense because we all know in winter the weather is very cold and freezing.</em>

b) Porque la temperatura de su cuerpo, siente el aire frio que entra por la ventana.

<em>I feel like this answer is the question, but it could also be an answer, sorry, I'm a little uncertain.</em>

c) Porque el calor de su cuerpo se propaga al medio ambiente, al ser la temperatura del niño mayor que la del aire exterior.

<em>This answer has nothing to do with the question, plus it is very false, our body heat is not enough to overcome the very cold temperature from outside.</em>

d) Porque la temperatura del aire es igual a la temperatura del cuerpo.

<em>This is false because again our body heat is not even compared to the freezing cold temperatures from the winter.</em>

<em />

<em />

<h2>Well, have a nice rest of the day!</h2><h3>ba baiii!</h3>
3 0
3 years ago
A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that incre
Evgen [1.6K]

Complete Question

A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 1.60 T in 0.99 s. What is the resulting induced current if the loop has a resistance of 1.20  \ \Omega

Answer:

The current is   I =  0.0007 41 \ A

Explanation:

From the question we are told that

   The  area is  A = 8.00 \ cm^2  = 8.0 *10^{-4} \  m^2

   The initial magnetic field at t_o = 0 \ seconds  is B_i = 0.500 \ T

   The magnetic field at t_1 = 0.99 \ seconds is  B_f  = 1.60 \ T

     The resistance is  R = 1.20  \ \Omega

Generally the induced emf is mathematically represented as

      \epsilon  =  A * \frac{B_f - B_i }{ t_f - t_o }

=>   \epsilon  =  8.0 *10^{-4} * \frac{1.60  - 0.500 }{ 0.99- 0  }

=>   \epsilon  = 0.000889 \ V

Generally the current induced is mathematically represented as

     I = \frac{\epsilon}{R }

=>  I = \frac{0.000889}{ 1.20 }  

=>  I =  0.0007 41 \ A  

6 0
3 years ago
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