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devlian [24]
3 years ago
7

Which SI units are combined to describe a force

Physics
1 answer:
Keith_Richards [23]3 years ago
4 0

Answer:

C.

Explanation:

kilograms and m/s^2

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Pls help 100 points plssssssss
natima [27]

Answer:

d

Explanation:

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2 years ago
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PLEASE HELP 40 POINTS
sergij07 [2.7K]

Liar, not even 40 points....

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A photon of wavelength 3.7 nm Compton scatters from an electron at an angle of 90°. What is the modified wavelength? (Enter your
viva [34]

Answer:

3.70242 nm

Explanation:

Using Compton effect formula

Δλ = ( h / mec) ( 1 - cosθ)

where h is planck constant = 6.62607 × 10 ⁻³⁴ m²kg/s

me, mass of an electron = 9.11 × 10⁻³¹ kg

c is the speed of light = 3 × 10⁸ m/s

Δλ = 6.62607 × 10 ⁻³⁴ m²kg/s / (9.11 × 10⁻³¹ kg × 3 × 10⁸ m/s ) ( 1 - cos 90°) =   0.242 × 10 ⁻¹¹ m = 2.42 × 10⁻¹² m  = 0.00242 nm

modified wavelength = 3.7 nm + 0.00242 nm = 3.70242 nm

7 0
4 years ago
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Why are we seeing extremely old light from Canopus instead of light in real-time?
Irina-Kira [14]

Answer:

Canopus is more than 300 light years away from earth. This means it takes the light we see more than 300 years to reach us.

8 0
3 years ago
Introduction: The specific heat capacity of a substance is the amount of energy needed to change the temperature of that substan
ki77a [65]

Answer:

A) 8,368 J

B) ) 0.893 J/gºC

Explanation:

A)

  • The heat gained by the water can be obtained solving the following equation:

       q_{g} = c_{w} * m *  \Delta T (1)

  • where cw = specific heat of water = 4.184 J/gºC
  • m= mass of water = 1,000 g
  • ΔT = 2ºC
  • Replacing these values in (1) we get:

       q_{g} = c_{w} * m *  \Delta T = 4.184 J/gºC*1,000 g* 2ºC = 8,368 J (2)

B)

  • Assuming that the heat energy gained by the water is equal to the one lost by the aluminum, we can use the same equation, taking into account that the energy is lost by the aluminum, so the sign is negative:  -8,368 J.
  • Replacing by the mass of aluminum (125 g), and the change in temperature (-74.95ºC), in (1), we can solve for the specific heat of aluminum, as follows:

       q_{l} = c_{Al} * m_{Al} *  \Delta T  (3)

⇒    -8,368 J = c_{Al}* 125 g * (-74.95ºC) (4)

       c_{Al} = \frac{-8,368J}{125g*(-74.95ºC} = 0.893 J/gºC (5)

  • which is pretty close to the Aluminum's accepted specific heat value of 0.900 J/gºC.

8 0
3 years ago
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