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insens350 [35]
1 year ago
15

Ancient Greek philosophers spent lots of time thinking about science and imaging explanations for the natural world. What part o

f the scientific process did they need to do
A. Testing


B. Creative thinking


C. Debate


D. Argumentation
Physics
1 answer:
Illusion [34]1 year ago
4 0

Answer:

Testing

Explanation:

Ancient Greek philosophers lived with the ideology to simply contemplate life. This means that their whole life revolved around thinking and questioning everything. This would include creative thinking, because they would sometimes come up with theories which require creativeness. They would often debate with their friends as to why their theory should be accepted or what their opinions were on the matter. More often than not, they argued a lot, and many philosophers went against some powerful people in the community and some were even sentenced to death.

The main process they didn't/couldn't do was the testing. They could never test certain theories because they did not have the means to.

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Solenoid 2 has twice the diameter, twice the length, and twice as many turns as solenoid 1. How does the field B2 at the center
klemol [59]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is  option 3

Explanation:

From the question we are told that

   The diameter of solenoid 1 is  d_1

   The length of solenoid 1 is   L_1

    The  number of turns of solenoid is  N_1

   The diameter of solenoid 2 is  d_2 = 2d_1

   The length of solenoid 2 is   L_2 = 2L_1

    The  number of turns of solenoid  2 is    N_2 = 2 N_1

Generally the magnetic in a solenoid is mathematically represented as

     B  =  \frac{\mu_o *  N  *  I }{L}

From this equation we see that

     B  \ \alpha \  \frac{N}{L}

     B   =  C   \frac{N}{L}

Here C stands for constant

=>   C =  \frac{B *  \frac{L}{N}

=>    \frac{B_1 *  \frac{L_1}{N_1}   = \frac{B_2 *  \frac{L_2}{N_2}

=>  \frac{B_1}{B_2 }  =  \frac{N_1 L _2}{ N_2L_1}

=>   \frac{B_1}{B_2 }  =  \frac{N_1 * (2 L_1)}{ (2 N_2)L_1}

=>   \frac{B_1}{B_2 }  =  1

=>   B_2 = B_1

4 0
2 years ago
If the blocks are released from rest, which way does the 10 kg block slide, and what is its acceleration? enter a positive value
anyanavicka [17]
<span>Let m1=10kg and m2=5kg and for our calculations assume right is positive and up is positive (note: for block hanging, the x axis is vertical so tilt your head to help) For m1 Sigma Fx = ma T - m1gsin35 = m1a where T = tension For m2 m2g - T = m2a Add equation together m1a + m2a = T-m1gsin35 + m2g - T a(m1 + m2) = m2g - m1gsin35 a= (5*9.8 - 10*9.8*sin35)/(10 + 5) a= -0.48m/s/s So the system is moving in the opposite direction of our set coordinate system where we said right positive, its negative so its moving left therefore down the ramp</span>
6 0
2 years ago
Read 2 more answers
Alex throws a 0.15-kg rubber ball down onto the floor. The ball’s speed just before impact is 6.5 m/s, and just after is 3.5 m/s
Jet001 [13]

Answer: Change in ball's momentum is 1.5 kg-m/s.

Explanation: It is given that,

Mass of the ball, m = 0.15 kg

Speed before the impact, u = 6.5 m/s

Speed after the impact, v = -3.5 m/s (as it will rebound)

We need to find the change in the magnitude of the ball's momentum. It is given by :

So, the change in the ball's momentum is 1.5 kg-m/s. Hence, this is the required solution.

Read more on Brainly.com - brainly.com/question/12946012#readmore

7 0
1 year ago
A moving 46.6 kg sled feels a 52.9 N friction force. what is the coefficient of friction
Setler [38]

Answer:

F=UR

52.9=U*46.6

U=52.9/46.6

U=1.135

4 0
2 years ago
Read 2 more answers
Show your work and resoning for the below requirement.
Leno4ka [110]

Answer:

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

Explanation:

We must work on this problem using the rotational equilibrium equations and then they compared the tension values that the cable supports.

Let's start with fixing a reference system on the hinge of the flag, we take as positive the anti-clockwise turn

 They indicate the weight of the pole W₁ = 120 lb and a length of L = 9 ft, the weight of the man W₂ = 150, we assume that the cable is at the tip of the pole

            - T_{y} L + W₂ L + W₁ L / 2 = 0

            T_{y} = W₂ + W₁ / 2

            T_{y} = 120 + 150/2

            T_{y} = 195 lb

we use trigonometry to find the cable tension

             sin 30 = T_{y} / T

             T = T_{y} / sin 30

             T = 195 / sin 30

             T = 390 lb

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

             T < 500 lb

4 0
2 years ago
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