Answer:
Sorry mate! I can't understand this language...
Answer:
54 days
Explanation:
We have to use the formula;
0.693/t1/2 =2.303/t log Ao/A
Where;
t1/2= half-life of phosphorus-32= 14.3 days
t= time taken for the activity to fall to 7.34% of its original value
Ao=initial activity of phosphorus-32
A= activity of phosphorus-32 after a time t
Note that;
A=0.0734Ao (the activity of the sample decreased to 7.34% of the activity of the original sample)
Substituting values;
0.693/14.3 = 2.303/t log Ao/0.0734Ao
0.693/14.3 = 2.303/t log 1/0.0734
0.693/14.3 = 2.6/t
0.048=2.6/t
t= 2.6/0.048
t= 54 days
Frequency.
The equation to find the velocity of a wave length is:
v=fλ
V stands for velocity
F stands for frequency
λ stands for wavelength
Answer:
stable isotopes have stable nuclei and do not show radioactivity, but for unstable isotopes it is the opposite
Explanation:
hope this helps, ask more questions if needed.
Answer:
603000 J
Explanation:
The following data were obtained from the question:
Energy required (Q) =...?
Mass (M) = 10000 g
Specific heat capacity (C) = 2.01 J/g°C
Overheating temperature (T2) = 121°C
Working temperature (T1) = 91°C
Change in temperature (ΔT) =.?
Change in temperature (ΔT) =T2 – T1
Change in temperature (ΔT) = 121 – 91
Change in temperature (ΔT) = 30°C
Finally, we shall determine the energe required to overheat the car as follow:
Q = MCΔT
Q = 10000 × 2.01 × 30
Q = 603000 J
Therefore, 603000 J of energy is required to overheat the car.
