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Doss [256]
3 years ago
6

What led kepler to abandon circular orbits and his discovery that planetary orbits are ellipses?

Physics
1 answer:
Nimfa-mama [501]3 years ago
4 0
<span>A deviation between observed and model values that was much larger than the observation uncertainties led Kepler to abandon circular orbits and his discovery that planetary orbits are ellipses</span>
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A system of mass 13 kg undergoes a process during which there is no work, the elevation decreases by 50 m, and the velocity incr
marin [14]

Answer:

The change in kinetic energy is 4.3875 kJ

The amount of energy transferred by heat for the process is -66.98 kJ

Explanation:

Given;

mass of the system, m = 13 kg

change in height, Δh = -50 m

initial velocity, u = 15 m/s

final velocity, v = 30 m/s

change in internal energy per mass, ΔU = -5 kJ/kg

The change in kinetic energy is given by;

ΔK.E = K.E₂ - K.E₁

ΔK.E = ¹/₂mv² - ¹/₂mu²

ΔK.E = ¹/₂m(v² - u²)

ΔK.E = ¹/₂ ₓ 13 (30² - 15²)

ΔK.E = 4387.5 J

ΔK.E = 4.3875 kJ

The amount of energy transferred by heat for the process;

Q = W + ΔP.E + ΔK.E + ΔU

Where;

ΔP.E = mgΔh

ΔP.E = 13 x 9.8 x (-50)

ΔP.E = -6370 J = -6.37 kJ

W = 0

ΔU = -5kJ/kg x 13kg

ΔU = -65 kJ

Q = W + ΔP.E + ΔK.E + ΔU

Q = 0 + (-6.37) + (4.3875) + (-65)

Q = -66.98 kJ

7 0
4 years ago
Would someone plz help a shister outttt
tiny-mole [99]

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8 0
4 years ago
State whether each of the following is a physical or chemical property
ikadub [295]

We need more information

7 0
3 years ago
All of the following are examples of mechanical energy except:
Paladinen [302]
The answer is B. The nuclei of elements.
8 0
3 years ago
Consider a standing wave in a one dimensional ideal medium of length "D" (like a vibrating string).
stira [4]

Answer:

a) 20 nodes    b) zero nodes

Explanation:

When we have standing waves in a bend we have nodes at the ends and the equation describes the number of possible waves in the string is

         L = n λ/2

Where λ is the wavelength, L is the length of the string, in our case it would be D and n is an entered. We can strip the wavelength of this expression

       

       λ = 2L / n

Let's calculate what value of n we have for a wavelength equal to D/10

       λ = 2D / n

       λ = D / 10

We match and calculate

       2D / n = D / 10

       2 / n = 1/10

       n = 20

Perform them for  λ = D / 20

       λ = 2D / n

       2D / n = D / 20

       n = 2 20 = 40

Since n is an inter there should be a wavelength for each value of n in the bone period there should be 20 different wavelengths

B) for La = 10D

       2D / n = 10D

       1 / n = 5

       n = 1/5 = 0.2

 

La = 20D

       2D / n = 20D

       1 / n = 10

      n = 1/10 = 0.1

These numbers are not entered so there can be no wave in this period

5 0
3 years ago
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