4 years ago

What is the definition of efficent

Physics

1 answer:

nika2105 [10]4 years ago

8 0

Efficient: the measure of how much energy is used in a process.

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The count rate of a radioactive source decreases from 1600 counts per minute to 400 counts per minute in 12 hours. What is the h

kirill115 [55]

Answer:

$t_{1/2}=6 h$

Explanation:

Let's use the decay equation.

$A=A_{0}e^{-\lambda t}$

Where:

A is the activity at t time
A₀ is the initial activity
λ is the decay constant

We know that $\lambda=\frac{ln(2)}{t_{1/2}}$

So we have:

$\lambda=\frac{ln(A/A_{0})}{t}$

$\frac{ln(2)}{t_{1/2}}=\frac{ln(A/A_{0})}{t}$

$t_{1/2}=\frac{t*ln(2)}{ln(A/A_{0})}$

$t_{1/2}=6 h$

Therefore, the half-life of the source is 6 hours.

I hope it helps you!

4 0

3 years ago

What is the correct definition of civics?

RSB [31]

Answer:

the definition of civics is the study of the rights and duties of citizenship.

so the answer is B.

hope this helps.

4 0

3 years ago

Read 2 more answers

calculate minimum number of incident photons per area and of the minimum dose needed to visualize an object of 1 mm squared usin

Lynna [10]

Answer:

minimum number of photon is 4.05 × $10^{7}$

Explanation:

given data

energy = 50 keV = 50 × $10^{3}$ eV = 50 × $10^{3}$ × 1.602× $10^{-19}$ J

thickness = 10^-3

contrast = 1%

to find out

number of incident photons

solution

we know here equation that is

E = n × h × ν .......................1

put here all these value

50 × $10^{3}$ = n × 6.6× $10^{-34}$ × c/ 1× $10^{-3}$

50 × $10^{3}$ × 1.602× $10^{-19}$ = n × 6.6× $10^{-34}$ ×( 3 × $10^{8}$ / 1× $10^{-3}$ )

solve it and find n

n = 4.05 × $10^{7}$

so here minimum number of photon is 4.05 × $10^{7}$

3 0

3 years ago

There are 15 water molecules in the air, and the relative

AlexFokin [52]

30 30 30 30 30 30 30 30 30 30

6 0

2 years ago

A particle (mass = 5.0 g, charge = 40 mC) moves in a region of space where the electric field is uniform and is given by Ex = –2

taurus [48]

Answer: 40m

Explanation: Distance is a function of speed/ velocity and time.

From the question the velocity along the z axis is given as 20 m/s at t = 0 secs We were ask to calculate the distance travelled when t = 2secs.

Recall that,

Velocity = distance travelled / time

Since we're calculating for distance we make it subject of formula;

Distance travelled = velocity * time

= 20m/s * 2s

= 40meter.

3 0

3 years ago