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Masja [62]
3 years ago
9

Sandy is on a road trip. She leaves at 8:00 AM. It takes her 2 hours to drive 200 kilometers. She stops at a rest stop for half

an hour. Then, she drives for 100 more kilometers, which takes her an hour and a half. What is Sandy's average velocity?
Physics
1 answer:
lutik1710 [3]3 years ago
0 0
The average velocity of Sandy is given by the total distance covered S divided by the total time taken t:
v= \frac{S}{t}

The total distance covered is
S=200 km+0+100 km=300 km
while the total time taken is 2 hours + half an hour (for the rest) + 1 hour and half, so
t=2h+0.5h+1.5 h=4 h
Therefore, the average velocity is 
v= \frac{S}{t}= \frac{300 km}{4 h}=75 km/h
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A hot-air balloon is rising upward with a constant speed of 2.03 m/s. When the balloon is 8.13 m above the ground, the balloonis
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A spherical balloon has a radius of 6.95 m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of a
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3 0
2 years ago
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)

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