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zaharov [31]
3 years ago
6

In the last part of the experiment you need to investigate the behaviour of gamma radiation as it passes through air. A gamma so

urce is placed at a distance d = 0.064 m away from the GM detector and the counts per minute recorded C = 585.
As we expect gamma radiation not to be absorbed by air an inverse square law type behaviour should be seen, where the count rate will decrease with distance. To see if this is correct you need to plot a graph, with data collected at different distances. Looking at the data given above, what point would you plot on the x-axis?
Physics
1 answer:
andrew11 [14]3 years ago
7 0

Answer:

linear cart C Vs 1/R²  or log cart C Vs R

Explanation:

Gamma radiation is very high energy electromagnetic rays, but its behavior is the same as for all radiation. By the principle of conservation of energy after the radiation is emitted, it must be distributed on a spherical surface which determines the behavior of the inverse of the square.

In this experiment you are measuring the rate of counts by time (C), this must be the dependent variable since it is not controlled by the experimenter and on the other hand it measures the distance (X) this is the independent variable since it is the one that we can control.

To make a graph with this data, the counting rate must be plotted against the inverse of the squared distance (1/R²). On the Y axis the counts per second and the X 1 / R² axis, with this graph a line must be obtained.

Another graph that we can make on double logarithmic paper where the Y axis plotted the counting rate and on the X axis the distance, the slope should give -2.

            C == A / R²

        Log C = log A -2 log R

With either of the two graphs, the law of the inverse of the square is tested

You might be interested in
How does natural selection produce change in a population of mice?
sergey [27]

Answer:

The population of the mice will decrease to the faster, stronger, and smarter mice because the weaker will die because of natural selection.

Explanation:

5 0
2 years ago
A cat dozes on a stationary merry-go-round, at a radius of 5.7 m from the center of the ride. Then the operator turns on the rid
tatyana61 [14]

Answer:

0.76

Explanation:

we are given:

radius (r) =5.7 m

speed (s) = 1 revolution in 5.5 seconds

acceleration due to gravity (g) = 9.8 m/s^{2}

coefficient of friction (Uk) = ?

 we can get the minimum coefficient of friction from the equation below

centrifugal force = frictional force

m x r x ω^{2} = Uk x m x g

r x ω^{2} = Uk x g

Uk = \frac{ r x ω^{2} }{g}

where ω (angular velocity) = \frac{2π}{time}

= \frac{2π }{5.5} = 1.14

Uk = \frac{ 5.7 x 1.14^{2} }{9.8} = 0.76

6 0
3 years ago
A 60 kg sprinter has a momentum of +600 kg-m/s when he crosses the finish
MakcuM [25]

Answer:

10 ms⁻¹

Explanation:

The amount of momentum that an object has is dependent upon two factors

  • mass of the moving object  
  • speed of motion

In terms of an equation,

Momentum (P) = Mass(m)×velocity(v)

                     P = m×v

                 600 = 60 × v ⇒ v = 10 ms⁻¹

3 0
3 years ago
A 5-kg ball collides inelastically head-on with a 10-kg ball, which is initially stationary. Which of the following statements i
NARA [144]

Answer:

The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball.

Explanation:

In inelastic collision, the total momentum is always conserved after collision but the kinetic energy is reduced.

Momentum is Mass X velocity.

5 kg ball is in motion, while 10 kg ball is stationary; that is zero velocity.

The momentum of 10 kg ball before collision is zero while the momentum of 5 kg ball before collision is more than zero. Therefore, the magnitude of change in momentum will not be equal.

Next possible options are in kinetic Energy

Initial Kinetic energy = \frac{1}{2}mu^2

Final kinetic energy =\frac{1}{2}mv^2

Change in kinetic energy = Final Kinetic Energy - Initial Kinetic Energy

Change in kinetic energy of 5kg ball = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2

Since the 5-kg ball has initial velocity (u), the magnitude of the change in velocity will be reduced.

Change in kinetic energy of 10kg ball:

the ball is initially at rest, therefore the initial velocity (u) will be zero (0)

Δ K.E = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2 = \frac{1}{2}m(v-0)^2 = \frac{1}{2}mv^2

From the solution above, the magnitude of the change in velocity experienced by 10 kg ball is higher than 5 kg ball.

Hence, The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball

4 0
3 years ago
Given a force of 100 N and an acceleration of 10 m/s2, what is the mass?
Drupady [299]
F=ma, so 100=m×10. Solve for m by dividing by 10. The mass is 10 kg.
5 0
2 years ago
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