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RoseWind [281]
3 years ago
10

Which two types of waves can transmit energy through a vacuum?

Physics
2 answers:
nalin [4]3 years ago
4 0
The answer would be the sound waves.
Bad White [126]3 years ago
3 0

<u><em>1.radio waves  </em></u>

2̶.̶s̶e̶i̶s̶m̶i̶c̶ ̶w̶a̶v̶e̶s

<em><u>3.sound waves</u></em><em>  </em>

4̶.̶w̶a̶t̶e̶r̶ ̶w̶a̶v̶e̶s

5̶.̶x̶-̶r̶a̶y̶s

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Another term for free enterprise system would be capitalism

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1.5m/s^2

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Guillaume puts a bottle of soft drink in a refrigerator and leaves it there until its temperature has dropped 15.1 K.
Zarrin [17]

Answer: (a) The magnitude of its temperature change in degrees Celsius is 15.1^{o}C.

(b) The magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is 27.2^{o}F.

Explanation:

(a)  Expression for change in temperature is as follows.

        |\Delta T| = |x - y|K

                         = 15.1 K

                    = |(x - 273.15) - (y - 273.15)|^{o}C

                    = |x - y|^{o}C

                    = 15.1^{o}C

Therefore, the magnitude of its temperature change in degrees Celsius is 15.1^{o}C.

(b)  Change in temperature from Celsius to Fahrenheit is as follows.

           F = 1.8C + 32

          C = \frac{F - 32}{1.8}

Since,   K = C + 273

or,    \Delta K = \Delta C = \frac{\Delta F}{1.8}

         \Delta F = 1.8 \Delta K

                      = 1.8 (15.1)

                      = 27.18^{o}F

or,                  = 27.2^{o}F

Thus, we can conclude that the magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is 27.2^{o}F.

7 0
3 years ago
Read 2 more answers
A pool ball moving 1.83 m/s strikes an identical ball at rest. Afterward, the first ball moves 1.15 m/s at a 23.3 degrees angle.
chubhunter [2.5K]

Answer:

 v_{1fy} = - 0.4549 m / s

Explanation:

This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved

initial. Before the crash

      p₀ = m v₁₀

final. After the crash

      p_{f} = m v_{1f} + m v_{2f}

Recall that velocities are a vector so it has x and y components

       p₀ = p_{f}

we write this equation for each axis

X axis

       m v₁₀ = m v_{1fx} + m v_{2fx}

       

Y Axis  

       0 = -m v_{1fy} + m v_{2fy}

the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components

      sin 23.3 = v_{2fy} / v_{2f}

      cos 23.3 = v_{2fx} / v_{2f}

      v_{2fy} = v_{2f} sin 23.3

      v_{2fx} = v_{2f} cos 23.3

we substitute in the momentum conservation equation

       m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3

       0 = - m v_{1f} sin θ + m v_{2f} sin 23.3

      1.83 = v_{1f} cos θ + 1.15 cos 23.3

       0 = - v_{1f} sin θ + 1.15 sin 23.3

      1.83 = v_{1f} cos θ + 1.0562

        0 = - v_{1f} sin θ + 0.4549

     v_{1f} sin θ = 0.4549

     v_{1f}  cos θ = -0.7738

we divide these two equations

      tan θ = - 0.5878

      θ = tan-1 (-0.5878)

       θ = -30.45º

we substitute in one of the two and find the final velocity of the incident ball

        v_{1f} cos (-30.45) = - 0.7738

        v_{1f} = -0.7738 / cos 30.45

        v_{1f} = -0.8976 m / s

the component and this speed is

       v_{1fy} = v1f sin θ

       v_{1fy} = 0.8976 sin (30.45)

       v_{1fy} = - 0.4549 m / s

8 0
3 years ago
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