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lara31 [8.8K]
3 years ago
12

How does half-life apply to real life?

Physics
2 answers:
Lyrx [107]3 years ago
6 0
The term half-life is defined as the time it takes for one-half of the atoms of a radioactive material to disintegrate."
wlad13 [49]3 years ago
3 0
half-life? what do you mean
You might be interested in
Determine the gravitational field 300km above the surface of the earth. How does this compare to the field on the earth's surfac
Serjik [45]
The strength of the gravitational field is given by:
g= \frac{GM}{r^2}
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.

In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
r=R+h=6370 km+300 km=6670 km= 6.67 \cdot 10^{6} m

And now we can use the previous equation to calculate the field strength at that altitude:
g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2}  = 8.95 m/s^2

And we can see this value is a bit less than the gravitational strength at the surface, which is g_s = 9.81 m/s^2.
4 0
3 years ago
An electric current in a metal consists of moving
Alik [6]
The answer would be (A) Protons
6 0
3 years ago
The radius of a planet is 2400 km, and the acceleration due to gravity at its surface is 3.6 m/s2.
kiruha [24]

Answer:

3.1\cdot10^{23}\:\mathrm{kg}

Explanation:

We can use Newton's Universal Law of Gravitation to solve this problem:

g_P=G\frac{m}{r^2}., where g_P is acceleration due to gravity at the planet's surface, G is gravitational constant 6.67\cdot 10^{-11}, m is the mass of the planet, and r is the radius of the planet.

Since acceleration due to gravity is given as m/s^2, our radius should be meters. Therefore, convert 2400 kilometers to meters:

2400\:\mathrm{km}=2,400,000\:\mathrm{m}.

Now plugging in our values, we get:

3.6=6.67\cdot10^{-11}\frac{m}{(2,400,000)^2},

Solving for m:

m=\frac{2,400,000^2\cdot3.6}{6.67\cdot 10^{-11}},\\m=\fbox{$3.1\cdot10^{23}\:\mathrm{kg}$}.

6 0
2 years ago
An electron moving parallel to a uniform electric field increases its speed from 2.0 × 10^7 m/s to 4.0 × 10^7 m/s over a distanc
julia-pushkina [17]

Answer:

E = 2.84 * 10^5 N/C

Explanation:

The speed increased from 2.0 * 10^7 m/s to 4.0 * 10^7 m/s over a 1.2 cm distance.

Let us find the acceleration:

v^2 = u^2 + 2as

(4.0 * 10^7)^2 = (2.0 * 10^7)^2 + 2 * a * 0.012\\\\(4.0 * 10^7)^2 - (2.0 * 10^7)^2 = 0.024a\\\\1.2 * 10^{15}= 0.024a\\\\a = 1.2 * 10^{15} / 0.024\\\\a = 5 * 10^{16} m/s^2

Electric force is given as the product of charge and electric field strength:

F = qE

where q = electric charge

E = Electric field strength

Force is generally given as:

F = ma

where m = mass

a = acceleration

Equating both:

ma = qE

E = ma / q

For an electron:

m = 9.11 × 10^{-31} kg

q = 1.602 × 10^{-19} C

Therefore, the electric field strength of the electron is:

E = \frac{9.11 * 10^{-31} * 5 * 10^{16}}{1.602 * 10^{-19}} \\\\E = 2.84 * 10^5 N/C

8 0
3 years ago
How does battery generate electric current in a circuit?
sveticcg [70]

Answer:

C. It creates negative electric charges and pushes them into the circuit

Explanation:

A battery has three main parts. The Cathode (Positive), the Electrolytes (energy source), and the Anode (negative). The anode takes energy from the electrolytes and creates more electrons. And because subatomic particles with the same charge don't like being in the same place, they move through the circuit to reach the cathode because opposites attract.

I hope this helps!

3 0
3 years ago
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