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umka2103 [35]
3 years ago
6

What are some other examples of scientific knowledge that must have resulted from indirect evidence?

Physics
1 answer:
n200080 [17]3 years ago
5 0

Answer:

The atomic model, the structure of the solar system, and the theory of black holes are examples of models created using indirect evidence.

Explanation:

I have no explanation sorry.

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Finish the following sentence. almost all of the oxygen (o2) one consumes in breathing is converted to _____________.
ludmilkaskok [199]
The answer is carbon dioxide (CO2)
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3 years ago
For questions 1-10 fill in the blank with the letter of the description that best matches the term
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B-Pitcher C-Catcher H-Strike I-Umpire G-Strike Zone E- Foul Ball F- Ball J- Pick-off D-Error A- Shortstop. I think (Sorry for them being out of order. I had to break them down)
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A particle with an initial linear momentum of 2.00 kg-m/s directed along the positive x-axis collides with a second particle, wh
ladessa [460]

Answer:

a) p₂ = 1.88 kg*m/s

   θ = 273.4 º

b)  Kf = 37% of Ko

Explanation:

a)

  • Assuming no external forces acting during the collision, total momentum must be conserved.
  • Since momentum is a vector, their components (projected along two axes perpendicular each other, x- and y- in this case) must be conserved too.
  • The initial momenta of both particles are directed one along the x-axis, and the other one along the y-axis.
  • So for the particle moving along the positive x-axis, we can write the following equations for its initial momentum:

       p_{o1x} = 2.00 kg*m/s (1)

       p_{o1y} = 0 (2)

  • We can do the same for the particle moving along the positive y-axis:

        p_{o2x} = 0 (3)

        p_{o2y} = 4.00 kg*m/s (4)

  • Now, we know the value of magnitude of the final momentum p1, and the angle that makes with the positive x-axis.
  • Applying the definition of cosine and sine of an angle, we can find the x- and y- components of the final momentum of the first particle, as follows:

       p_{f1x} = 3.00 kg*m/s * cos 45 = 2.12 kg*m/s (5)

      p_{f1y} = 3.00 kg*m/s sin 45 = 2.12 kg*m/s  (6)

  • Now, the total initial momentum, along these directions, must be equal to the total final momentum.
  • We can write the equation for the x- axis as follows:

       p_{o1x} + p_{o2x} = p_{f1x} + p_{f2x}  (7)

  • We know from (3) that p₀₂ₓ = 0, and we have the values of p₀1ₓ from (1) and pf₁ₓ from (5) so we can solve (7) for pf₂ₓ, as follows:

       p_{f2x} = p_{o1x} - p_{f1x} = 2.00kg*m*/s - 2.12 kg*m/s = -0.12 kg*m/s (8)

  • Now, we can repeat exactly the same process for the y- axis, as follows:

       p_{o1y} + p_{o2y} = p_{f1y} + p_{f2y}  (9)

  • We know from (2) that p₀1y = 0, and we have the values of p₀₂y from (4) and pf₁y from (6) so we can solve (9) for pf₂y, as follows:

       p_{f2y} = p_{o1y} - p_{f1y} = 4.00kg*m*/s - 2.12 kg*m/s = 1.88 kg*m/s (10)

  • Since we have the x- and y- components of the final momentum of  the second particle, we can find its magnitude applying the Pythagorean Theorem, as follows:

       p_{f2} = \sqrt{p_{f2x} ^{2} + p_{f2y} ^{2} }  = \sqrt{(-0.12m/s)^{2} +(1.88m/s)^{2}} = 1.88 kg*m/s (11)

  • We can find the angle that this vector makes with the positive x- axis, applying the definition of tangent of an angle, as follows:

       tg \theta = \frac{p_{2fy} }{p_{2fx} } = \frac{1.88m/s}{(-0.12m/s} = -15.7 (12)

  • The angle that we are looking for is just the arc tg of (12) which measured in a counter-clockwise direction from the positive x- axis, is just 273.4º.

b)

  • Assuming that both masses are equal each other, we find that the momenta are proportional to the speeds, so we find that the relationship from the final kinetic energy and the initial one can be expressed as follows:

       \frac{K_{f}}{K_{o} } = \frac{v_{f1}^{2} + v_{f2} ^{2}}{v_{o1}^{2} + v_{o2} ^{2} } = \frac{12.5}{20} = 0.63 (13)

  • So, the final kinetic energy has lost a 37% of the initial one.

6 0
2 years ago
A Brayton cycle has air into the compressor at 95 kPa, 290 K, and has an efficiency of 50%. The exhaust temperature is 675 K. Fi
motikmotik

Answer:

The specific heat addition is 773.1 kJ/kg

Explanation:

from table A.5 we get the properties of air:

k=specific heat ratio=1.4

cp=specific heat at constant pressure=1.004 kJ/kg*K

We calculate the pressure range of the Brayton cycle, as follows

n=1-(1/(P2/P1)^(k-1)/k))

where n=thermal efficiency=0.5. Clearing P2/P1 and replacing values:

P2/P1=(1/0.5)^(1.4/0.4)=11.31

the temperature of the air at state 2 is equal to:

P2/P1=(T2/T1)^(k/k-1)

where T1 is the temperature of the air enters the compressor. Clearing T2

11.31=(T2/290)^(1.4/(1.4-1))

T2=580K

The temperature of the air at state 3 is equal to:

P2/P1=(T3/T4)^(k/(k-1))

11.31=(T3/675)^(1.4/(1.4-1))

T3=1350K

The specific heat addition is equal to:

q=Cp*(T3-T2)=1.004*(1350-580)=773.1 kJ/kg

3 0
3 years ago
How much kinetic energy will an electron gain if it accelerates through a potential difference of 23,000 volts in a cathode ray
Sauron [17]

Answer:

3.68x10-15 J

Explanation:

I hope this is the right answer sorry if im wrong

8 0
2 years ago
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