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umka2103 [35]
2 years ago
6

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

Physics
1 answer:
n200080 [17]2 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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A 10.0 cm3 sample of copper has a mass of 89.6
Romashka-Z-Leto [24]
Density is mass divides by volume, so
89.6g / 10cm^3 =8.96g /cm^3

*cm^3 is a standard unit of volume*
4 0
2 years ago
Calculate the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s.
Aleonysh [2.5K]

De broglie wavelength, \lambda = \frac{h}{mv}, where h is the Planck's constant,  m is the mass and v is the velocity.

h = 6.63*10^{-34}

Mass of hydrogen atom,  m = 1.67*10^{-27}kg

v = 440 m/s

Substituting

   Wavelength \lambda = \frac{h}{mv} = \frac{6.63*10^{-34}}{1.67*10^{-27}*440} = 0.902 *10^{-9}m = 902 *10^{-12}m

1 pm = 10^{-12}m\\ \\ So , \lambda =902 pm

So  the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s is 902 pm

7 0
2 years ago
Steam undergoes an adiabatic expansion in a piston–cylinder assembly from 100 bar, 360°C to 1 bar, 160°C. What is work in kJ per
vfiekz [6]

Answer:

work is 130.5 kJ/kg

entropy change is 1.655 kJ/kg-k

maximum  theoretical work is 689.4 kJ/kg

Explanation:

piston cylinder assembly

100 bar, 360°C to 1 bar, 160°C

to find out

work  and amount of entropy  and magnitude

solution

first we calculate work i.e heat transfer - work =   specific internal energy @1 bar, 160°C  - specific internal energy @ 100 bar, 360°C    .................1

so first we get some value from steam table with the help of 100 bar @360°C and  1 bar @ 160°C

specific volume = 0.0233 m³/kg

specific enthalpy = 2961 kJ/kg

specific internal energy = 2728 kJ/kg

specific entropy = 6.004 kJ/kg-k

and respectively

specific volume = 1.9838 m³/kg

specific enthalpy = 2795.8 kJ/kg

specific internal energy = 2597.5 kJ/kg

specific entropy = 7.659 kJ/kg-k

now from equation 1 we know heat transfer q = 0

so - w =   specific internal energy @1 bar, 160°C  - specific internal energy @ 100 bar, 360°C

work = 2728 - 2597.5

work is 130.5 kJ/kg

and entropy change formula is i.e.

entropy change =  specific entropy ( 100 bar @360°C)  - specific entropy ( 1 bar @160°C )

put these value we get

entropy change =  7.659 - 6.004

entropy change is 1.655 kJ/kg-k

and we know maximum  theoretical work = isentropic work

from steam table we know specific internal energy is 2038.3 kJ/kg

maximum  theoretical work = specific internal energy - 2038.3

maximum  theoretical work = 2728 - 2038.3

maximum  theoretical work is 689.4 kJ/kg

3 0
2 years ago
A proton starts from rest and gains 8.35 x 10^-14 joule of kinetic energy as it accelerates between points A and B in an electri
antoniya [11.8K]

Answer:

5.22 x 10^5 V

Explanation:

guessed on castle learning and got it right

6 0
2 years ago
Read 2 more answers
thin films of soap sometime display an array of colours. this display is the result of: a) reflection, diffraction, and interfer
kumpel [21]
The array of colors observed when viewing films of soap are caused by the interference, and from the reflections of light from two nearby surfaces (outer surface and inner surface). Among the choices, the closest answer would be B. reflection, refraction, and interference.
7 0
2 years ago
Read 2 more answers
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