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3 years ago

What happends when you slowly pull out the paper while theirs a coin on top

Physics

2 answers:

IgorC [24]3 years ago

7 0

The correct answer is “ the coin will stay on the same place on the paper and will move along with the paper “

HACTEHA [7]3 years ago

3 0

The coin would probably slowly slide down whichever end it is closer to, weigh down the end of the paper, and fall off. It kinda depends on how you hold the paper though, because if it is directly in middle, and you hold it on both ends, it will stay on the paper.

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Help with these please someone?

DIA [1.3K]

Answer:

8. 2.75·10^-4 s^-1

9. No, too much of the carbon-14 would have decayed for radiation to be detected.

Explanation:

8. The half-life of 42 minutes is 2520 seconds, so you have ...

1/2 = e^(-λt) = e^(-(2520 s)λ)

ln(1/2) = -(2520 s)λ

-ln(1/2)/(2520 s) = λ ≈ 2.75×10^-4 s^-1

___

9. Reference material on carbon-14 dating suggests the method is not useful for time periods greater than about 50,000 years. The half-life of C-14 is about 5730 years, so at 65 million years, about ...

6.5·10^7/5.73·10^3 ≈ 11344

half-lives will have passed. Whatever carbon 14 may have existed at the time will have decayed completely to nothing after that many half-lives.

7 0

3 years ago

How are mixtures and pure substances related?

Elden [556K]

You are talking about make sure's and pearl substance I thought you was talking about mix in with something

4 0

3 years ago

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N

Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

$r=\sqrt{(r_{1})^2+(r_{2})^2}$

Put the value into the formula

$r=\sqrt{(9.8)^2+(2.1)^2}$

$r=10.02\ cm$

We need to calculate the force

Using formula of force

$F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}$

Force F₁₂,

$F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}$

$F_{12}=13.33\ N$

$F_{21}=-13.33\ N$

Force F₂₃,

$F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}$

We need to calculate the value of third charge

$F_{net}=F_{21}+F_{23}$

$4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}$

$q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}$

$q_{3}=7.99\times10^{-7}\ C$

$q_{3}=0.8\times10^{-6}\ C$

Hence, The value of third charge is 0.8μC.

4 0

3 years ago

An amusement park ride consists of a rotating circular platform 8.26 m in diameter from which 10 kg seats are suspended at the e

VashaNatasha [74]

To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.

PART A) We will begin by finding the two net distances.

$r = \frac{8.26}{2} = 4.13m$