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A spacecraft orbits the Earth at height of 1600km. Calculate the escape velocity for the spacecraft.

guapka [62]

Answer:

what do u need help with

Explanation:

Calculate the escape velocity for the spacecraft. [G= 6.67×10^-11Nm^2kg^-2, mass of the Earth= 5.97×10^24kg, radius of the Earth= ...

7 0

2 years ago

A bike, a truck, and a train—all without passengers, motors, or engines—roll down the same hill. Put the vehicles in order from

Pepsi [2]

Answer: WAIT WHATTTT i have that same test due today and the answer is in explanation

Explanation:

Bike, truck train. we are in the same school i think. Its imma say the incisal JMES Im Lusi i used to help in the library

4 0

3 years ago

Why are special techniques and extinguishing agents are required to fight combustible metals fires?

d1i1m1o1n [39]

Cucucicicicivigovogovphphpbpb

7 0

3 years ago

A group of atoms with aligned magnetic poles are known as which of the following?

cestrela7 [59]

A magnetic domain is a group of atoms aligns with magnetic poles. Domains are usually <span>light and dark stripes visible within each grain.</span>

4 0

3 years ago

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A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri

GREYUIT [131]

Answer:

$|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))$

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

$\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r$

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)$

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

$\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}$

Now, substitute this into 'dF':

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)$

Now we can integrate dF over the rod.

$\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)$

4 0

3 years ago