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Yuki888 [10]
3 years ago
7

Nuclear waste disposal is one of the largest issues with nuclear power. Cesium-137 is one of the high level waste products in an

uranium nuclear fission reaction. Safely disposing of Cs-137 is extremely important due both to this isotope's long half-life, 30.07 years, and it ability to substitute for potassium in biological reactions. A 5.2 mg sample of Cs-137 requires half-lives to decay to 0.65 mg. After 150 years, of Cs-137 remains. After 6 half-lives, of Cs-137 remains. After years, 0.0102 mg of Cs-137 remains. Finally, after approximately 601 years - half-lives, only of the Cs-137 remains.
Physics
1 answer:
vodomira [7]3 years ago
7 0

Answer:

A sample of 5.2 mg  decays to .65 mg or to 1/8 of its original amount.

1/8 = 1/2 * 1/2 * 1/2 or 3 half-lives.

3 * 30.07 = 90 yrs for 5.2 mg to decay to .65 mg

You can get these other numbers similarly:

5.2 / .0102 = 510  requires about 9  half-lives which is 30 * 9 = 270 yrs

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A football wide receiver rushes 16 m straight down the playing field in 2.9 s (in the positive direction). He is then hit and pu
Bumek [7]

Answer:

a) v1 = 5.52m/s

b) v2 = -1.52m/s

c) v3 = 4.62m/s

d) vt = 3.85m/s

Explanation:

The velocity of the football wide receiver is his displacement per unit time.

Velocity v = (displacement d)/time t

v = d/t .....1

For each of the cases, equation 1 would be used to calculate the velocity.

a) v1 = d1/t1

d1= 16m

t1 = 2.9s

v1 = 16m/2.9s

v1 = 5.52m/s

b) v2 = d2/t2

d2 = -2.5m

t2 = 1.65s

v2 = -2.5/1.65

v2 = -1.52m/s

c) v3 = d3/t3

d3 = 24m

t3 = 5.2s

v3 = 24/5.2

v3 = 4.62m/s

d) vt = dt/tt

dt = 16m - 2.5m + 24m = 37.5m

tt = 2.9 + 1.65 + 5.2 = 9.75s

vt = 37.5/9.75

vt = 3.85m/s

5 0
3 years ago
Car A hits car B (initially at rest and of equal mass) from behind while going 15 m/s Immediately after the collision, car B mov
Mamont248 [21]

Given :

Initial speed of car A is 15 m/s and initial speed of car B is zero.

Final speed of car A is zero and final speed of car B is 10 m/s.

To Find :

What fraction of the initial kinetic energy is lost in the collision.

Solution :

Initial kinetic energy is :

K.E_i = \dfrac{15^2m}{2} + 0\\\\K.E_i = \dfrac{225 m}{2}

Final kinetic energy is :

K.E_f = \dfrac{10^2m}{2} + 0\\\\K.E_f = \dfrac{100m}{2}

Now, fraction of initial kinetic energy loss is :

Loss = \dfrac{\dfrac{225m}{2}-\dfrac{100m}{2}}{\dfrac{100m}{2}}\\\\Loss = \dfrac{125}{100}\\\\Loss = 1.25

Therefore, fraction of initial kinetic energy loss in the collision is 1.25 .

6 0
2 years ago
A weight lifter does 586 J of work on a weight that he lifts in 3.5 seconds. What is the power with which he lifts the weight?
Neko [114]

Answer:

Explanation:

The quantity of energy transferred by a force when it is applied to a body and causes that body to move in the direction of the force work.

7 0
2 years ago
Assume an axon has an internal diameter of 1μm and a myelin sheath 1μm thick. The internal specific resistance is 100 Ω cm. For
SpyIntel [72]

Answer:

1.27\times 10^{12}\Omega/m

Explanation:

We are given that

Diameter=d=\mu m

Thickness=1\mu m

Radius=r=\frac{d}{2}=\frac{1}{2}\mu m=0.5\times 10^{-6} m

Using 1\mu m=10^{-6} m

Dielectric constant=8

Resistance =R=2\times 10^5\Omega cm^2

Internal specific resistance=r=100 ohm cm=100\times \frac{1}{100}\Omega-m=1\Omega m

Using 1 m=100 cm

Internal resistance per unit length=\frac{r}{A}=\frac{1}{\pi r^2}=\frac{1}{3.14\times (0.5\times 10^{-6})^2}=1.27\times 10^{12}\Omega/m

Using \pi=3.14

Internal resistance per unit length=1.27\times 10^{12}\Omega/m

8 0
3 years ago
What is nuclear fission? why is this process important to stars?
KIM [24]

Nuclear fusion is a reaction in which two or more atomic nuclei come very close and then collide at a very high speed and join to form a new nucleus. This process is important to stars because they get their energy from the nuclear fusion process

4 0
3 years ago
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