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Vadim26 [7]
2 years ago
6

Suppose that placing 0.3 inch of lead in front of a gamma source reduces the count rate from 1045 cps to 573 cps. What is um^-1

in g/cm^2?
Physics
1 answer:
Ainat [17]2 years ago
5 0

Answer:

14.49 g/cm²

Explanation:

I = Io e^-(ux)

Where:

I = 573

Io = 1045

x = 0.3 inches and

rho = 11.4g/cm^3

Using the conversion constant

1 inch = 2.54 cm;

0.3 inches = 0.3 * 2.54 cm

0.3 inches = 0.762 cm

I/Io = e^-(ux), or say

Io/I = e^(ux), taking the In of both sides

ln(Io/I) = ux, making u subject of formula

u = 1/x * ln(Io/I)

u = 1/0.762 * ln(1045/573)

u = 1.312 * 0.6

u = 0.787

Next, we say that

u/rho = 0.7872/11.4 = 0.069

And finally, we make

1/(u/rho) to be our final answer

Inverse of the answer is = 14.49 g/cm²

Therefore, the um^-1 in g/cm^2? is 14.49

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Answer:

μsmín = 0.1

Explanation:

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       F_{frmax} = \mu_{s} *F_{n} (1)

       where  μs is the coefficient of static friction, and Fn is the normal force,

       perpendicular to the wall and aiming to the center of rotation.

  • This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
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       F_{c} =  m* \omega^{2} * r (2)

       where ω is the angular velocity of the riders, and r the distance to the

      center of rotation (the  radius of the circle), and m the mass of the

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     F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)

  • When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

       m* g = m* \mu_{smin} * \omega^{2} * r (4)

  • (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
  • Cancelling the masses on both sides of (4), we get:

       g = \mu_{smin} * \omega^{2} * r (5)

  • Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

      60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)

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2 years ago
This time particle A starts from rest and accelerates to the right at 65.5 cm/s
FrozenT [24]

Answer:

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Explanation:

As we know that the particle A starts from Rest with constant acceleration

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d = 0 + \frac{1}{2}(65.5)t^2

d_1 = 32.75 t^2 cm

Now we know that B moves with constant speed so in the same time B will move to another distance

d_2 = 44 \times t

now we know that B is already 349 cm down the track

so if A and B will meet after time "t"

then in that case

d_1 = 349 + d_2

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on solving above kinematics equation we have

t = 4 s

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