Respuesta :
Answer:
The answers as per the given problem is solved in the segment below.
Explanation:
According to the question,
(a)
[tex]F=ILB[/tex]
By putting the values, we get
[tex]=10(2\times 10^4)(5\times 10^{-5})[/tex]
[tex]=10 \ N[/tex]
(b)
[tex]W=Fx[/tex]
or,
[tex]=Fvt[/tex]
[tex]=10(7.80\times 10^3)(3600)[/tex]
[tex]=2.81\times 10^8 \ J[/tex]
(c)
[tex]\frac{1}{2}mv'^2 =\frac{1}{2}mv^2-\omega[/tex]
or,
[tex]v=[v^2-\frac{2 \omega}{m} ]^\frac{1}{2}[/tex]
[tex]=[(7.8\times 10^3)^2-\frac{2(2.81\times 10^8)}{1\times 10^5} ]^{\frac{1}{2} }[/tex]
[tex]=7.79\times 10^3 \ m/s[/tex]
now,
[tex]\Delta v=v-v'[/tex]
[tex]=0.36 \ m/s[/tex]
(d)
- The speed difference is around 1 percent throughout a week's trip. Generally, a decrease throughout the speed will indeed lead the orbit towards spiral inwards, ever since the speed would not have been adequate to maintain that rotational velocity.
- As nothing more than a result, this same shuttle could very well could use a little more propellant to sustain the velocity of aspiration if the orbit were ever to spiral somewhat within the vehicle.
