7. PE=0.5×700n/m×0.9m^2
0.9^2=0.81m
0.5×700×0.81= 283.5J
8. 2000=0.5×(x)×1.5m^2
1.5^2= 0.25
0.25×0.5=0.125
2000=0.125 (x)
2000/0.125=x
x=16000 n/m
9. 4000=0.5 (375 n/m)×(x)^2
0.5×187.5 (x)
4000/187.5=21.3333333333
Answer:
The smallest part of a millimeter that can be read with a digital caliper with a four digit display is 0.02mm. Thus, it has to be converted to centimetre. So, divide by 10, we then have 0.02/10= *0.002cm* not mm.
Answer:
"8 units" is the appropriate answer.
Explanation:
According to the question,
Throughout equilibrium all particles are of equivalent intensity, and as such the integrated platform's total energy has been uniformly divided across all individuals.
Now,
The total energy will be:
= 
= 
The total number of particles will be:
= 
= 
hence,
Energy of each A particle or each B particle will be:
= 
= 
Answer:
Darwin's theory of natural selection lacked an adequate account of ... Darwinian principles now play a greater role in biology than ever before, .... Sadly, even if Mendel had lived to see the rediscovery of his work, he probably .... evolutionary forces are acting, a genetic equivalent to Galileo's law of inertia.
Explanation:
