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enyata [817]
2 years ago
11

What is the half life-life of your 100 atoms of Carbon-14?

Physics
1 answer:
Naddika [18.5K]2 years ago
3 0

Answer:

Answer: It takes 5,730 years for half the carbon-14 to change to nitrogen; this is the half-life of carbon-14. After another 5,730 years only one-quarter of the original carbon-14 will remain

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If you measure the amount of work accomplished in a particular time interval, u have measured-
serious [3.7K]

Answer:

The power

Explanation:

We know that the work definition is given by the following expression:

W = F * d

where:

F = force [Newtons] [N]

d = distance [meters] [m]

W = work [Joules]

And the expression that defines the work done by unit of time is called - <u>Power</u>, therefore:

P = W/t

where:

P = power [watts] [w]

W = work [Joules] [J]

t = time [seconds] [s]

6 0
3 years ago
g n diffraction, the formula for minima is given by a times s i n (theta )equals m lambda, where a is the width of the slit, the
ki77a [65]

Answer:

θ = 22.2

Explanation:

This is a diffraction exercise

        a sin θ = m λ

The extension of the third zero is requested (m = 3)

They indicate the wavelength  λ = 630 nm = 630 10⁻⁹ m and the width of the slit  a = 5 10⁻⁶ m

         sin  θ = m λ / a

         sin  θ = 3 630 10⁻⁹ / 5 10⁻⁶

         sin  θ = 3.78 10⁻¹ = 0.378

          θ = sin⁻¹  0.378

         

to better see the result let's find the angle in radians

          θ = 0.3876 rad

let's reduce to degrees

         θ = 0.3876 rad (180º /π rad)

         θ = 22.2º

4 0
3 years ago
Help me and you will meet your favorite singer at your door.
Goshia [24]
Sunlight, because it provides a source of energy is the answer because plants also provide a food source (please put as brainliest answer)
4 0
3 years ago
Read 2 more answers
Calculate the angle θ between the radius-vector of the point and the positive x axis (measured counterclockwise from the positiv
Y_Kistochka [10]

The point obviously is in the 3rs quadrant

So

စ= tan^-1( y/x)-180

စ= -89.7°

6 0
3 years ago
A 1.0-kg block of aluminum is at a temperature of 50°C. How much thermal energy will it lose when its temperature is reduced by
Marat540 [252]

Answer:

22425 J

Explanation:

From the question,

Applying

Q = cm(t₂-t₁).................. Equation 1

Where Q = Thermal Energy, c = specific heat capacity of aluminium, m = mass of aluminium, t₂ = Final Temperature, t₁ = Initial Temperature.

Given: c = 897 J/kg.K, m = 1.0 kg, t₁ = 50 °C, t₂ = 25 °C (The final temperature is reduced by half)

Substitute these values into equation 1

Q = 897×1×(25-50)

Q = 897×(-25)

Q = -22425 J

Hence the thermal energy lost by the aluminium is 22425 J

5 0
3 years ago
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