It starts or speeds up.
Some substances need certain amount of energy in order to react with each other.
Answer:
V4=9.197v
Explanation:
Given:
V1= 18v ,V2= 12v ,r1=r5=58ohms ,r2=r6=124ohms , r3=47ohms ,r4= 125ohms
V4= I4R4 = V2/(R4 + R5)×R4
V4= 12×125 /(125 + 58)
V4=1500/183 =9.197v
Answer:
The mass of the ice block is equal to 70.15 kg
Explanation:
The data for this exercise are as follows:
F=90 N
insignificant friction force
x=13 m
t=4.5 s
m=?
applying the equation of rectilinear motion we have:
x = xo + vot + at^2/2
where xo = initial distance =0
vo=initial velocity = 0
a is the acceleration
therefore the equation is:
x = at^2/2
Clearing a:
a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2
we use Newton's second law to calculate the mass of the ice block:
F=ma
m=F/a = 90/1.283=70.15 kg
Answer:
c. 2 MeV.
Explanation:
The computation of the binding energy is shown below
![= [Zm_p + (A - Z)m_n - N]c^2\\\\=[(1) (1.007825u) + (2 - 1 ) ( 1.008665 u) - 2.014102 u]c^2\\\\= (0.002388u)c^2\\\\= (.002388) (931.5 MeV)\\\\=2.22 MeV](https://tex.z-dn.net/?f=%3D%20%5BZm_p%20%2B%20%28A%20-%20Z%29m_n%20-%20N%5Dc%5E2%5C%5C%5C%5C%3D%5B%281%29%20%281.007825u%29%20%2B%20%282%20-%201%20%29%20%28%201.008665%20u%29%20-%202.014102%20u%5Dc%5E2%5C%5C%5C%5C%3D%20%280.002388u%29c%5E2%5C%5C%5C%5C%3D%20%28.002388%29%20%28931.5%20MeV%29%5C%5C%5C%5C%3D2.22%20MeV)
= 2 MeV
As 1 MeV = (1 u) c^2
hence, the binding energy is 2 MeV
Therefore the correct option is c.
We simply applied the above formula so that the correct binding energy could come
And, the same is to be considered
<span>Answer: "a cold front" .
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