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Natasha2012 [34]
3 years ago
11

A person standing on the edge of a high cliff throws a rock straight up with an initial velocity, v0v0, of 13.0m/s13.0m/s. on th

e way back down the rock misses the edge of the cliff and continues to fall. ignoring the effects of air resistance, calculate the position and velocity of the rock a
Physics
1 answer:
Firdavs [7]3 years ago
8 0
For the position, we can use the equation below for our calculations:
H = h + v₀²/2g
where 
H is the total height from the base of the cliff to the highest point of the rock on air
h is the height of the cliff
v₀ is the initial height
g is equal to 9.81 m/s²

H = h + 13²/2(9.81)
H = h + 8.61 m

Because the height of the cliff is not known, the answer would just be h +8.61 m.

For the velocity, we use the equation: v = √2gH. Substituting the answer for H, the velocity is

v = √2(9.81)(h+8.61)
v = 4.43√(h+8.61)
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Dimension equation of work
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Jon's bathtub is rectangular and its base is 18 ft2. (a) How fast is the water level rising if Jon is filling the tub at a rate
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Compute the expected shell-model quadrupole moment of 209Bi () and compare with the experimental value, - 0.37 b
Over [174]

Answer:

0.22 b

Explanation:

Quadrupole moment of the nucleon is,

Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2}

And also,

R^{2}=R^{2} _{0}A^{\frac{2}{3} }

And, R _{0}=1.2\times 10^{-15}m

Now,

Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2} _{0}A^{\frac{2}{3} }

For Bismuth j=\frac{9}{2} and A is 209.

Q=-\frac{2\frac{9}{2} -1}{2(\frac{9}{2} +1)}\frac{3}{5}(1.2\times 10^{-15}) ^{2}(209)^{\frac{2}{3} }\\Q=0.628\times 35.28\times 10^{-30} \\Q=22.15\times 10^{-30} m^{2} \\Q=0.2215\times 10^{-28} m^{2} \\Q=0.22 barn

Therefore, the expected value of quadrupole is 0.22 b which is quite related with experimental value which is 0.37 b

3 0
3 years ago
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