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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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An atom emission of light with a specific amount of energy confirms that ?

marusya05 [52]

The correct answer is B, electrons emit and absorb energy based on their position around the nucleus.

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3 years ago

Recall that photosynthetic rates remain relatively constant in regions near the equator. Imagine that tropical environments pers

solniwko [45]

Answer:E.

a straight line sloping upward (atmospheric CO2 levels would not seasonally oscillate, but would have increased over time)

Explanation: Photosynthesis is the process through which green plants and certain other organisms make their food using carbon dioxide and water in the he presence of Sunlight.

During photosynthesis, energy conversion from light energy into chemical energy takes place.

Photosynthesis is one of the major processes that helps to eliminate and reduce the quantity of carbon dioxide in the atmosphere,it also serve to make oxygen present in the atmosphere.

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2 years ago

Suppose your surface body temperature averaged 90 degrees F. How much radiant energy in W/m^2 would be emitted from your body?

Debora [2.8K]

$493 \; \text{W}\cdot \text{m}^{-2}$ .

<h3>Explanation</h3>

The Stefan-Boltzmann Law gives the energy radiation <em>per unit area</em> of a black body:

$\dfrac{P}{A} = \sigma \cdot T^{4}$

where,

$P$ the total power emitted,
$A$ the surface area of the body,
$\sigma$ the Stefan-Boltzmann Constant, and
$T$ the temperature of the body in degrees Kelvins.

$\sigma = 5.67 \times 10^{-8} \;\text{W}\cdot \text{m}^{-2} \cdot \text{K}^{-4}$ .

$T = 90 \; \textdegree{}\text{F} = (\dfrac{5}{9} \cdot (90-32) + 273.15) \; \text{K} = 305.372 \; \text{K}$ .

$\dfrac{P}{A} = \sigma \cdot T^{4} = 5.67 \times 10^{-8} \times 305.372^{4} = 493\; \text{W}\cdot \text{m}^{-2}$ .

Keep as many significant figures in $T$ as possible. The error will be large when $T$ is raised to the power of four. Also, the real value will be much smaller than $493\; \text{W}\cdot \text{m}^{-2}$ since the emittance of a human body is much smaller than assumed.

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3 years ago

An astronaut exploring a distant solar system lands on an unnamed planet with a radius of 2530 km. When the astronaut jumps upwa

Natali [406]

Answer:

1.38*10^18 kg

Explanation:

According to the Newton's law of universal gravitation:

$F=G*\frac{m_a*m_p}{r^2}$

where:

G= Gravitational constant (6.674×10−11 N · (m/kg)2)

ma= mass of the astronaut

mp= mass of the planet

$F=m_a.a\\(v_f )^2=(v_o)^2+2.a.\Delta y\\\\a=\frac{(v_f)^2-(v_o)^2}{2.\Delta y}\\\\a=\frac{(0)^2-(4.29m/s)^2}{2.0.64m}=14.38m/s^2\\\\F=m_a*14.38m/s^2$

so:

$m_a*14.38m/s^2=(6.674*10^{-11}N.(m/kg)^2)*\frac{m_a.m_p}{(2.530*10^3m)^2}\\m_p=\frac{14.38m/s^2(2.530*10^3m)^2}{(6.674*10^{-11}N.(m/kg)^2)}\\\\m_p=1.38*10^{18}kg$