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3 years ago

There is a 50 g sample of Ra-229. It has a half-life of 4 minutes.

Physics

1 answer:

Sloan [31]3 years ago

4 0

Via half-life equation we have:

$A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h} }$

Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes. By plugging those values in we get:

$A_{final}=50(\frac{1}{2})^\frac{12}{4}=50(\frac{1}{2})^{3}=50(\frac{1}{8})=6.25g$

There is 6.25 grams left of Ra-229 after 12 minutes.

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Dos cargas puntuales q1 = −50μC y q2 = +30μC se encuentran

alina1380 [7]

Answer:

Datos:

q1 = -50 μC = $50*10^{-6}$

q2 = +30 μC = $30*10^{-6}$

F = 10 N

a) x si la <em>F = 10N</em>

Aplicando la Ley de Coulomb:

x = $\sqrt\frac{k0 * q1 *q2}{F}$ = $\sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{10}$ = 1,162m

b) x si la <em>F = 20 N</em>

x=<em> </em> $\sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{20}$ <em> </em>= 0,822m

c)x si la <em>F = 50 N</em>

x = $\sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{50}$ = 0,520m

3 0

3 years ago

Why are you able to observe the Doppler effect on earth with sound waves but not with light waves?

VladimirAG [237]

Answer: When an ambulance passes with its siren blaring, you hear the pitch of the siren change: as it approaches, the siren’s pitch sounds higher than when it is moving away from you. This change is a common physical demonstration of the Doppler effect.

Explanation:

6 0

3 years ago

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent

azamat

The wavelengths of the constituent travelling waves CANNOT be 400 cm.

The given parameters:

<em>Length of the string, L = 100 cm</em>

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The wavelengths of the constituent travelling waves is calculated as follows;

$L = \frac{n \lambda}{2} \\\\n\lambda = 2L\\\\\lambda = \frac{2L}{n}$

for first mode: n = 1

$\lambda = \frac{2\times 100 \ cm}{1} \\\\\lambda = 200 \ cm$

for second mode: n = 2

$\lambda = \frac{2L}{2} = L = 100 \ cm$

For the third mode: n = 3

$\lambda = \frac{2L}{3} \\\\\lambda = \frac{2 \times 100}{3} = 67 \ cm$

For fourth mode: n = 4

$\lambda = \frac{2L}{4} \\\\\lambda = \frac{2 \times 100}{4} = 50 \ cm$

Thus, we can conclude that, the wavelengths of the constituent travelling waves CANNOT be 400 cm.

The complete question is below:

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:

A. 400 cm

B. 200 cm

C. 100 cm

D. 67 cm

E. 50 cm

Learn more about wavelengths of travelling waves here: brainly.com/question/19249186

5 0

2 years ago

A turntable with a moment of inertia of 7.2 × − ⋅ rotates freely with an angular speed of 6.5 ⁄ . Riding on the rim of the turnt

irina1246 [14]

Answer:

turntable with a moment of inertia of 7.2 × − ⋅ rotates freely with an angular speed of 6.5 ⁄ . Riding on the rim of the turntable, 2 from the center, is a hamster. When the hamster walks to the center of the turntable, the angular speed of the turntable becomes ⁄. What is the mass of hamster?

Explanation:

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2 years ago

1- A train travels 92 kilometers in 3 hours, and then 55 kilometers in 3 hours. What is its average speed?

Hunter-Best [27]

<h3>Answer:</h3>

24.5 km/h
4 17/27 m/s
11/3 m/s²

<h3>Explanation:</h3>

1. The average speed is the ratio of total distance to total time:

... speed = distance/time = (92 km +55 km)/(3 h +3h) = (147 km)/(6 h)

... = 24.5 km/h

2. speed = distance/time = (125 m)/(27 s) = 4 17/27 m/s

3. a = ∆v/∆t = (15 m/s -4 m/s)/(3 s) = 11/3 m/s²

8 0

3 years ago