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Anarel [89]
3 years ago
11

A diver initially moving horizontally with speed v dives dives offthe edge of a vertical cliff and lands in the water a distance

dfrom the base of the cliff. How far from the base of the cliffwould the diver have landed if the diver initially had been movinghorizontally with speed 2v?
A. d
B. √(2d)
C. 2d
D. 4d
E. It cannot be determined unless the hieght of the cliff isknown
Physics
1 answer:
Aliun [14]3 years ago
7 0

Answer:

Option C is correct

Explanation:

"The time is determined by the vertical distance. The formula is sqr(2d/a) = t. There still is no acceleration in the horizontal direction."

For the first drive

d = d

t = sqr(2d/a)

r = v  

For the Second drive

d = ??

t = sqr(2d/a)

r = 2v

Since the times are same, equate the results.

t = d/v = d1/2v

v*d1 = 2v*d The v's cancel because they are related.

d1 = 2d.

You go twice as far as you did before.

Option C is correct

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Calculate the separation between the two lowest levels for an O2 molecule in a one-dimensional container of length 5.0 cm. At wh
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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
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