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3 years ago

How can we use radioactive sources?

Physics

2 answers:

docker41 [41]3 years ago

5 0

Answer:

Alpha - smoke detectors

Beta - tracers (medical imaging), monitoring the thickness of materials (industry - production of plastics etc.)

Gamma - treating cancer, sterilising medical equipment

DedPeter [7]3 years ago

4 0

Answer:

Today, to benefit humankind, radiation is used in medicine, academics, and industry, as well as for generating electricity. In addition, radiation has useful applications in such areas as agriculture, archaeology (carbon dating), space exploration, law enforcement, geology (including mining), and many others.

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Question number 8 Plz help

Dennis_Churaev [7]

To me, that sounds like the "Law of Conservation of Energy".

5 0

2 years ago

A potential difference of 107 mV exists between the inner and outer surfaces of a cell membrane. The inner surface is negative r

sergij07 [2.7K]

Answer:

The workdone is $W = 1.712 *10^{-20 } \ J$

Explanation:

From the question we are told that

The potential difference is $V = 107 mV = 107 *10^{-3} \ V$

Generally the charge on $Na^{+}$ is $Q_{Na^{+}} = 1.60 *10^{-19 } \ C$

Generally the workdone is mathematically represented as

$W = Q_{Na^{+}}V$

=> $W = 1.60 *10^{-19 } * 107 *10^{-3}$

=> $W = 1.712 *10^{-20 } \ J$

8 0

2 years ago

ALL OF MY POINTS FOR THIS!

scoray [572]

Newton's 2nd law of motion:

   Force = (mass) x (acceleration)

   = (1,127 kg) x (6 m/s² forward)

   = (1,127 x 6) newtons forward

   = 6,762 newtons forward
______________________________

   Momentum = (mass) x (speed)

   =   (69 kg) x (6 m/s)

   = 414 kg-m/s

3 0

3 years ago

Why is it helpful to study the light that comes off stars?

asambeis [7]

It is helpful to study the light that comes off stars because C. The light from a star gives hints of what elements make up a star.

6 0

3 years ago

Find the resistance of wire of 0.65m Radius 0.25 and resistivity 3x10-6 OHM

densk [106]

Complete Question:

Find the resistance of a wire of length 0.65 m, radius 0.25 mm and resistivity 3 * 10^{-6} ohm-metre.

Answer:

Resistance = 9.95 Ohms

Explanation:

Given the following data;

Length = 0.65 m

Radius = 0.25 mm to meters = 0.00025 m

Resistivity = 3 * 10^{-6} ohm-metre.

To find the resistance of the wire;

Mathematically, resistance is given by the formula;

$Resistance = P \frac {L}{A}$

Where;

P is the resistivity of the material.
L is the length of the material.
A is the cross-sectional area of the material.

First of all, we would find the cross-sectional area of the wire.

Area of circle = πr²

Substituting into the equation, we have;

Area = 3.142 * (0.00025)²

Area = 3.142 * 6.25 * 10^{-8}

Area = 1.96 * 10^{-7} m²

Now, to find the resistance of the wire;

$Resistance = 3 * 10^{-6} * \frac {0.65}{1.96 * 10^{-7}}$

$Resistance = 3 * 10^{-6} * 3316326.531$

Resistance = 9.95 Ohms

5 0

2 years ago