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2 years ago

The concentration of an isotope is 1/500 of its initial value after one day. What is the rate of decay?

1 answer:

8 0

The exponential growth/decay formula is given by A = Pe^(rt), where A is the final amount, P is the initial amount, r is the rate of growth/decay and t is time.

We are given A = (1/500)P and t=1, so substituting:
(1/500)P = Pe^(r*1)
Cancelling out P on both sides:
1/500 = e^r
ln(1/500) = ln(e^r)
r = -6.215

Rate of decay is -6.215 per day.

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Fill in the blanks to complete each statement about the heating of Earth's surface.

cricket20 [7]

Answer:

absorption and insolation.

Explanation:

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3 years ago

A gas of helium atoms at 273 k is in a cubical container with 25.0 cm on a side. (a) what is the minimum uncertainty in momentum

qwelly [4]

wave function of a particle with mass m is given by ψ(x)={ Acosαx −

2α

≤x≤+

2α

0 otherwise , where α=1.00×1010/m.

(a) Find the normalization constant.

(b) Find the probability that the particle can be found on the interval 0≤x≤0.5×10−10m.

(d) Find its average momentum.

(e) Find its average kinetic energy −0.5×10−10m≤x≤+0.5×10−10m.

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3 years ago

When is thermal equilibrium achieved between two objects?

ratelena [41]

C) When both objects have the same temperature.

Hope this helps!

5 0

3 years ago

Read 2 more answers

The law of suggests that the orbit of planets is not circular but .

SVEN [57.7K]

One of Kepler's laws is that the orbits of planets are elliptical. It's not a suggestion. BTW, circles are ellipses too, but so special that their likelihood is close to zero.

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3 years ago

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Your office has a 0.025 m^3 cylindrical container of drinking water. The radius of the container is about 13 cm. Required:a. Whe

uranmaximum [27]

Answer:

a) 105935.7 Pa

b) 103630.35 Pa

Explanation:

The volume of the container = 0.025 m^3

The radius of the container = 13 cm = 0.13 m

We have to find the height of the tank

From the equation for finding the volume of the cylinder,

V = $\pi r^2h$

where

V is the volume of the cylinder

h is the height of the cylinder

substituting values, we have

0.025 = 3.142 x $0.13^2$ x h

0.025 = 0.0531h

h = 0.025/0.0531 = 0.47 m

Pressure at the bottom of the tank P = ρgh

where

ρ is the density of water = 1000 kg/m^3

g is the acceleration due to gravity = 9.81 m/s^2

h is the depth of water which is equal to the height of the tank

substituting values, we have

P = 1000 x 9.81 x 0.47 = 4610.7 Pa

atmospheric pressure = 101325 Pa

therefore, the pressure in the tank bottom above atmospheric pressure = 101325 Pa + 4610.7 Pa = 105935.7 Pa

b) For half way down the container, depth of water will be = 0.47/2 = 0.235 m

pressure P = 1000 x 9.81 x 0.235 = 2305.35 Pa

This pressure above atmospheric pressure = 101325 Pa + 2305.35 Pa = 103630.35 Pa

7 0

3 years ago