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Kruka [31]
2 years ago
12

True or False: The height from which cliff divers jump affects the velocity at which they will fall. Velocity is speed in a cert

ain direction. Include your reasoning with your answer.
Physics
1 answer:
Juli2301 [7.4K]2 years ago
8 0

Answer:

the diver's speed is independent of the launch height.

Explanation:

For this exercise we must use Newton's second law

              fr - W = m a

the friction force has the general form

              fr = b v

Let's analyze this equation to find out what happens with the speed of the distant club.

When jumping, the initial speed is zero, so the friction force is zero and has an acceleration equal to the acceleration of gravity, as the speed increases the friction force increases decreasing the acceleration until it becomes zero, when it arrives at this value the velocity it has is called terminal velocity and this velocity remains fixed in relation to the trajectory.

             fr = W

             v = cte

The distance or time in which this equilibrium is reached is relatively fast, so the diver's speed is independent of the launch height.

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Chất rắn có tính dị hướng là vật rắn
kari74 [83]

Answer:

Anisotropy, in physics, the quality of exhibiting properties with different values when measured along axes in different directions. Anisotropy is most easily observed in single crystals of solid elements or compounds, in which atoms, ions, or molecules are arranged in regular lattices.

Explanation:

HOPE IT HELPS

6 0
2 years ago
Calculate the separation between the two lowest levels for an O2 molecule in a one-dimensional container of length 5.0 cm. At wh
MariettaO [177]

Answer:

The separation between the two lowest levels =  1.24 * 10^{-39}J

The values of n where the energy of molecule reaches 1/2 kT at 300K = 2.2 * 10^{9}

The separation at this level = 1.8 * 10^{-30}J

Explanation:

Knowing the formula

En = \frac{n^{2} h^{2}  }{8 mL^{2} }

Mass of oxygen molecule

m (O2) = 32 amu * \frac{1.6605 * 10^{-27 kg} }{1 amu}

So the energy diference between the two lowest levels:

E2 - E1 = \frac{3h^{2} }{8mL^{2} }

E2 - E1 =  \frac{3 * (6.626 * 10^{-34} Js)^{2} } {8 * 32 amu * (\frac{1.6605 * 10^{-27 kg} }{1 amu})* (5*10^{-2})^{2}   } = 1.24 * 10^{-39}J

Now we should find n where the energy of molecule reaches 1/2 kT

En = \frac{n^{2} h^{2}  }{8 mL^{2} } = \frac{1}{2}kT

\frac{h^{2}  }{8 mL^{2} } = 4.13 * 10^{-14}J

n^{2} *  (4.13 * 10^{-14}J) = \frac{1}{2} (1.38 * 10^{-23}JK^{-1}) * 300K

n = 2.2 * 10^{9}

by the end is necessary to calculate the separation of the level

En - En-1 = (n^{2} - (n - 1)^{2}) * \frac{h^{2}  }{8 mL^{2} }

              = 1.8 * 10^{-30}J

4 0
3 years ago
An electron is released from rest at a distance of 0.300 m from a large insulating sheet of charge that has uniform surface char
jarptica [38.1K]
Yes that is the correct answer
8 0
3 years ago
The position of a 2.2 kg mass is given by x=(2t3â5t2) m, where t is in seconds.
Juliette [100K]
B b b b b b b bb  bb   bb b b b b b b b b b 
6 0
3 years ago
Mr. Smith used 606 kWh for the month. If the service charge was $61.37 (see back page of the bill), what is the approximate char
Murrr4er [49]

Answer:

.10/KWh

Explanation:

divide 606 by 61.37 and you get .1012...

7 0
3 years ago
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