Explanation:
the answer to question one is complete the answer to question to is from negative to positive
Answer:
V=15.3 m/s
Explanation:
To solve this problem, we have to use the energy conservation theorem:
![U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}](https://tex.z-dn.net/?f=U_e%2BK_i%2BU_%7Bgi%7D%2BW_%7Bfriction%7D%3DK_f%2BU_%7Bgf%7D)
the elastic potencial energy is given by:
![U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J](https://tex.z-dn.net/?f=U_e%3D%5Cfrac%7B1%7D%7B2%7D%2Ak%2Ax%5E2%5C%5CU_e%3D%5Cfrac%7B1%7D%7B2%7D%2A1100N%2Fm%2A%284m%29%5E2%5C%5CU_e%3D8800J)
The work is defined as:
![W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J](https://tex.z-dn.net/?f=W_%7Bfriction%7D%3DF_f%2Ad%2Acos%28%5Ctheta%29%5C%5CW_%7Bfriction%7D%3D40N%2A2.5m%2Acos%28180%29%5C%5CW_%7Bfriction%7D%3D-100J)
this work is negative because is opposite to the movement.
The gravitational potencial energy at 2.5 m aboves is given by:
![U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J](https://tex.z-dn.net/?f=U_%7Bgf%7D%3Dm%2Ag%2Ah%5C%5CU_%7Bgf%7D%3D60kg%2A9.8%2A2.5%5C%5CU_%7Bgf%7D%3D1470J)
the gravitational potential energy at the ground and the kinetic energy at the begining are 0.
![8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s](https://tex.z-dn.net/?f=8800J%2B0%2B0%2B-100J%3D%5Cfrac%7B1%7D%7B2%7D%2A62kg%2Av%5E2%2B1470J%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B2%288800J-100J-1470J%29%7D%7B62kg%7D%7D%5C%5Cv%3D15.3m%2Fs)
Answer:
Temperature=3.86*10^7K
V1=
1.4
Explanation:
Deuterium (2H), when heated to sufficiently high tempera- ture, undergoes a nuclear fusion reaction that results in the production of helium. The reaction proceeds rapidly at a temperature, T, at which the average kinetic energy of the deuterium atoms is 8 × 10216 J. (At this temperature, deu- terium molecules dissociate completely into deuterium atoms.)
(a) Calculate T in kelvins (atomic mass of 2H 5 2.015).
(b) For the fusion reaction to occur with ordinary H atoms, the average energy of the atoms must be about 32 × 10216 J. By what factor does the average speed of the 1H atoms differ from that of the 2H atoms of part (a)?
k.E =3/2kT
K.E is the kinetic energy
k=boltzmann constant
T=Temperature in kelvin
juxtaposing the given values we have
T=2(KE)/3k
T=2*8*10^-16/(3*1.38*10^-23)
T=3.86*10^7K
Temperature=3.86*10^7K
b. average speed V=![\sqrt{8RT/\pi }M](https://tex.z-dn.net/?f=%5Csqrt%7B8RT%2F%5Cpi%20%7DM)
R is the gas constant
T absolute temperature in kelvin
M molar mass
since R.T,PI are constant
we are comparing between hydrogen and helium particles
V=![\alpha \sqrt{1/M}](https://tex.z-dn.net/?f=%5Calpha%20%5Csqrt%7B1%2FM%7D)
V1/V2=![\sqrt{M2/M1}](https://tex.z-dn.net/?f=%5Csqrt%7BM2%2FM1%7D)
V1=V2![\sqrt{2/1}](https://tex.z-dn.net/?f=%5Csqrt%7B2%2F1%7D)
V1=V2* 1.4
V1=
1.4
The characteristic of stars that primarily determines a star's color
is temperature. The correct
answer between all the choices given is the third choice. I am hoping that this
answer has satisfied your query and it will be able to help you in your
endeavor, and if you would like, feel free to ask another question.