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Olin [163]
3 years ago
14

What is the kinetic energy of a 620.0 kg roller coaster moving with a velocity of 9.00 m/s?

Physics
2 answers:
Alex73 [517]3 years ago
6 0

Answer:

The kinetic energy is equal to 25,110 J.

Explanation:

Darina [25.2K]3 years ago
3 0
The kinetic energy of a moving object is calculated through the equation,
                         KE = 0.5mv²
where m is the mass and v is the velocity. 
Substituting the known values,
                         KE = 0.5(620 kg)(9 m/s)²
                           KE = 25,110 J
Thus, the kinetic energy is equal to 25,110 J. 
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They compress or expand depending on amount of pressure or depending on the temperature
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2 years ago
Read 2 more answers
Calculate the force on an object with mass of 50kg and gravity of 10​
GarryVolchara [31]
Answer: 500 N

Explanation:

The formula to find the force exerted by a mass, we may use F = mg, where g, the gravity, and a, the acceleration, can be interchangeable in the formula.

1) F = 50 x 10
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Hope this helps, brainliest would be appreciated :)
6 0
2 years ago
b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul
Nat2105 [25]

Answer:

b) Betelgeuse would be \approx 1.43 \cdot 10^{6} times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is \approx 1.37 \cdot 10^{-5} } the statement saying that it would be like a second Sun is incorrect

Explanation:

The start brightness is related to it luminosity thought the following equation:

B = \displaystyle{\frac{L}{4\pi d^2}} (1)

where B is the brightness, L is the star luminosity and d, the distance from the star to the point where the brightness is calculated (measured). Thus:

b) B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}} and B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}} where L_{Sun} is the Sun luminosity (3.9 x 10^{26} W) but we don't need to know this value for solving the problem. ly is light years.

Finding the ratio between the two brightness we get:

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is 1.581 \cdot 10^{-5}\ ly. Then

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.

4 0
3 years ago
Mac and Keena are experimenting with pulses on a rope. Mac vibrates one end up and down while Keena holds the other end. This cr
Mariulka [41]
Mac and Keena are experimenting with pulses on a rope. Mac vibrates one end up and down while Keena holds the other end. This creates a pulse which they observe moving from end to end. How does the position of a point on the rope before the start of the pulse compare to its position after the pulse passes? Explain your reasoning.
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Three-fourths of the elements on the
77julia77 [94]

Answer:

b

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