Describe radioactive decay.

1 answer:

7 0

The spontaneous transformation of an unstable atomic nucleus into a lighter one, in which radiation is released in the form of alpha particles, beta particles, gamma rays, and other particles. The rate of decay of radioactive substances such as carbon 14 or uranium is measured in terms of their half-life .

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A Ford Taurus driven 12,000 miles a year will use about 650 gal of gasoline compared to a Ford Explorer that would use 850 gal.

irga5000 [103]

Answer:

19 800 lbm of carbon dioxide more.

Explanation:

<u>Taurus</u>

Amount of gasoline used in 5 years = 650 ga/year * 5 years = 3250 ga

amount of carbon dioxide released = 19.8 lbm/ga * 3250 ga = 64 350 lbm

<u>Explorer</u>

Amount of gasoline used in 5 years = 850 ga/year * 5 years = 4250 ga

amount of carbon dioxide released = 19.8 lbm/ga * 4250 ga = 84 150 lbm

Extra amount of CO2 released = 84 150 lbm - 64 350 lbm = 19 800 lbm

4 0

4 years ago

A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C

mario62 [17]

Answer:

3.62 m and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}$

$\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}$

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}$

$\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}$

x = - 1.4 m

6 0

4 years ago

What is the slant of a line on a graph called

tatiyna

Ashleybotello0129,

What is the slant of a line on a graph called "the slope." Slopes are in math, and science. Slopes look like slanted lines start from one side and to another.

Hope this helps!