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3 years ago

6

When the amount of a radioactive substance is reduced by 50%, what has occurred? a half-life beta radiation alpha radiation gamm

a radiation?

2 answers:

melomori [17]3 years ago

6 0

The answer is half life, I just did a test that had that on it

Vinil7 [7]3 years ago

5 0

This is the half life of the substance. The half life is the time it takes for the substance to become reduced by half as a consequence of decay.

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What is the formula for the speed of a car

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Does altitude has an effect on weight? PLEASE HELP!

Studentka2010 [4]

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3 years ago

A physicist's right eye is presbyopic (i.e., farsighted). This eye can see clearly only beyond a distance of 97 cm, which makes

Ivan

Answer:

f = 19,877 cm and P = 5D

Explanation:

This is a lens focal length exercise, which must be solved with the optical constructor equation

1 / f = 1 / p + 1 / q

where f is the focal length, p is the distance to the object and q is the distance to the image.

In this case the object is placed p = 25 cm from the eye, to be able to see it clearly the image must be at q = 97 cm from the eye

let's calculate

1 / f = 1/97 + 1/25

1 / f = 0.05

f = 19,877 cm

the power of a lens is defined by the inverse of the focal length in meters

P = 1 / f

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3 years ago

A police car is traveling north on a straight road at a constant 16.0 m/s. An SUV traveling north at 30.0 m/s passes the police

Nastasia [14]

Answer:

It will take 15.55s for the police car to pass the SUV

Explanation:

We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:

1. $x=x_{0}+vt$

2. $x=x_{0}+v_{0}t+\frac{at^{2}}{2}$

Since both cars will travel the same distance x, we can equal both formulas and solve for t:

$vt = v_{0}t+\frac{at^2}{2}\\\\ 16\frac{m}{s}t =30\frac{m}{s}t-\frac{1.8\frac{m}{s^{2}} t^{2}}{2}$

We simplify the fraction present and rearrange for our formula so that it equals 0:

$0.9\frac{m}{s^{2}} t^{2}-14\frac{m}{s}t=0 \\\\ t(0.9\frac{m}{s^{2}}t-14\frac{m}{s})=0$

In the very last step we factored a common factor t. There is two possible solutions to the equation at $t=0$ and:

$0.9\frac{m}{s^{2}}t-14\frac{m}{s}=0 \\\\ 0.9\frac{m}{s^{2}}t =14\frac{m}{s} \\\\ t =\frac{14\frac{m}{s}}{0.9\frac{m}{s^{2}}}=15.56s$

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at $t=0 s$ (when the SUV passed the police car) and $t=15.56s$ (when the police car catches up to the SUV)

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3 years ago