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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

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What is the level of incidence
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Answer:

The incidence rate is typically expressed as the number of cases per person-year of observation. Only new cases are considered when computing the incidence rate, while cases that were diagnosed earlier are excluded. The “population at risk” measure is usually obtained from census data.

Explanation:

The incidence rate is typically expressed as the number of cases per person-year of observation. Only new cases are considered when computing the incidence rate, while cases that were diagnosed earlier are excluded. The “population at risk” measure is usually obtained from census data.

5 0
3 years ago
A Honda Civic travels in a straight line along a road. Its distance x from a stop sign is given as a function of time t by the e
andrew-mc [135]
Average velocity = (x( 2.08 ) - x ( 0 )) / ( 2.08 s - 0 s )
x ( 2.08 ) = 1.42 * 2.08² - 0.05 * 2.08³ =
= 1.42 * 4.3264 - 0.443456 = 6.143484 - 0.443456 ≈ 5.7 m
v = ( 5.7 m - 0 m) / (2.08 s - 0 s ) = 5.7 / 2.08 m/s = 27.4 m/s
3 0
3 years ago
Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.
shusha [124]

Answer:

F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

                                  r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2  }

- Then compute the angle that Force makes with the y axis:

                                 cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

                                 F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

                                F_net = F_1,2 + kq / y^2

- Hence,

                                F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

- Now for the limit y >>a:

                              F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2}  )

- Insert limit i.e a/y = 0

                              F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2})  \\\\F_n = 3*k*q/y^2

Hence the Electric Field is off a point charge of magnitude 3q.

8 0
3 years ago
A car with a total mass of 1800 kg (including passengers) is driving down a washboard road with bumps spaced 4.9 m apart. The ri
Drupady [299]

Answer:

k = 9.6 x 10^5 N/m or 9.6 kN/m

Explanation:

First, we need to use the expression to calculate the spring constant which is:

w² = k/m

Solving for k:

k = w²*m

To get the angular velocity:

w = 2πf

The problem is giving the linear velocity of the car which is 5.7 m/s. With this we can calculate the frequency of the car:

f = V/x

f = 5.7 / 4.9 = 1.16 Hz

Now the angular velocity:

w = 2π*1.16

w = 7.29 rad/s

Finally, solving for k:

k = (7.29)² * 1800

k = 95,659.38 N/m

In two significant figures it'll ve 9.6 kN/m

5 0
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2, coal is a renewable resource
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