Answer:
<h2><em>
6000 counts per second</em></h2>
Explanation:
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;
2000 counts per second = 1 meter ... 1
In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;
x count per second = 3 meter ... 2
Solving the two expressions simultaneously for x we will have;
2000 counts per second = 1 meter
x counts per second = 3 meter
Cross multiply to get x
2000 * 3 = 1* x
6000 = x
<em></em>
<em>This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample</em>
It is diffraction
Explanation:
The opening is the aperture
D = distance between the cars at the start of time = 680 km
v₁ = speed of one car
v₂ = speed of other car = v₁ - 10
t = time taken to meet = 4 h
distance traveled by one car in time "t" + distance traveled by other car in time "t" = D
v₁ t + v₂ t = D
(v₁ + v₂) t = D
inserting the values
(v₁ + v₁ - 10) (4) = 680
v₁ = 90 km/h
rate of slower car is given as
v₂ = v₁ - 10
v₂ = 90 - 10 = 80 km/h