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>
Answer:
700 mL or 0.0007 m³
Explanation:
P₁ = Initial pressure = 2 atm
V₁ = Initial volume = 350 mL
P₂ = Final pressure = 1 atm
V₂ = Final volume
Here the temperature remains constant. So, Boyle's law can be applied here.
P₁V₁ = P₂V₂

So, volume of this sample of gas at standard atmospheric pressure would be 700 mL or 0.0007 m³
Answer:
The specific heat capacity is q_{L}=126.12kJ/kg
The efficiency of the temperature is n_{TH}=0.67
Explanation:
The p-v diagram illustration is in the attachment
T_{H} means high temperature
T_{L} means low temperature
The energy equation :
= R*
in(
/
)



The specific heat capacity:
=q_{h}*(T_{L}/T_{H})
q_{L}=378.36 * (400/1200)
q_{L}=378.36 * 0.333
q_{L}=126.12kJ/kg
The efficiency of the temperature will be:
=1 - (
/
)
n_{TH}=1-(400/1200)
n_{TH}=1-0.333
n_{TH}=0.67
Answer:
0.0021576N
Explanation:
F=(k)(q1q2/r^2)
F=(8.99×10^9)(3×10^-6)(2×10^-6)/(5^2)
F=0.0021576N
"the field of force surrounding a body of finite mass in which anotherbody would experience an attractive force that is proportional to theproduct of the masses and inversely proportional to the square of thedistance between <span>them."
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