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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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The force of attraction between a ball is F=.........×10^-¹¹
DIA [1.3K]

Answer:

4.45×10¯¹¹ N

Explanation:

From the question given above, the following data were obtained:

Mass of ball (M₁) = 4 Kg

Mass of bowling pin (M₂) = 1.5 Kg

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Distance apart (r) = 3 m

Force of attraction (F) =?

The force of attraction between the ball and the bowling pin can be obtained as follow:

F = GM₁M₂ / r²

F = 6.67×10¯¹¹ × 4 × 1.5 / 3²

F = 4.002×10¯¹⁰ / 9

F = 4.45×10¯¹¹ N

Therefore, the force of attraction between the ball and the bowling pin is 4.45×10¯¹¹ N

8 0
3 years ago
What evidence did wegener have to support his theory of plate tectonics?
PSYCHO15rus [73]

Answer:

Wegener first thought of this idea by noticing that the different large landmasses of the Earth almost fit together like a jigsaw puzzle. The continental shelf of the Americas fits closely to Africa and Europe, and other continents showed the same trend. Wegner also analyzed both sides of the Atlantic Ocean for rock type, geological structures and fossils and noticed that there was a significant similarity between matching sides of the continents, especially in fossil plants.

8 0
3 years ago
HELPPPPPPP PLEASE!!! does anybody have the answer key to edge. physics lab report : mechanical equivalent to heat? one page of d
Karo-lina-s [1.5K]

Answer:

the mechanical equivalent of heat states that motion and heat are mutually interchangeable and that in every case

Explanation:

6 0
3 years ago
Read 2 more answers
A car which is traveling at a velocity of 15 m/s undergoes an acceleration of 6.5 m/s2 over a distance of 340 m. How fast is it
fomenos

Answer:68.15m/s

Explanation:

<u><em>Given: </em></u>

v₁=15m/s

a=6.5m/s²

v₁=?

x=340m

<u><em>Formula:</em></u>

v₁²=v₁²+2a (x)

<u>Set up:</u>

=\sqrt{15m/s} ^{2} +2(6.5m/s^2)(340m)

<h2><u><em>Solution:</em></u></h2><h2><u><em>68.15m/s</em></u></h2>

<u />

6 0
3 years ago
A child of mass 40.0 kg is in a roller coaster car that travels in a loop of radius 7.00 m. at point a the speed of the car is 1
pav-90 [236]
I attached the missing picture.
The force of seat acting on the child is a reaction the force of child pressing down on the seat. This is the third Newton's law. The force of a child pressing down the seat and the force of the seat pushing up on the child are the same.
There two forces acting on the child. The first one is the gravitational force and the second one is centrifugal force. In this example, the force of gravity is always pulling down, but centrifugal force always acts away from the center of circular motion.
Part A
For point A we have:
F_a=F_cf-F_g
In this case, the forces are aligned, centrifugal is pointing up and gravitational is pulling down.
F_a=m\frac{v^2}{r}-mg=179 $N
Part B
At the point, B situation is a bit more complicated. In this case force of gravity and centrifugal force are not aligned. We have to look at y components of this forces, y-axis, in this case, is just pointing upward.
F=F_{cf}\cos(30)-mg=m\frac{v^2}{r}\cos(30)-mg=153.2$N
Part C
The child will stay in place at point A when centrifugal force and force of gravity are in balance:
F_g=F_{cf}\\&#10;mg=m\frac{v^2}{r}\\&#10;gr=v^2\\&#10;v=\sqrt{gr}=8.29\frac{m}{s}

6 0
3 years ago
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