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3 years ago

How long does a car battery last with lights

1 answer:

4 0

The answer is 4 and 7 hours .<span>In terms of how long until you can't start the car, a suppose or assuming the car battery is fully charged probably in about <span>45 to 90 minutes.</span></span>

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He user of a machine applies force to the machine over the _____ distance.

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The blank distance is your answer

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2 years ago

Question 3 Please, i need help! Thank you

Tom [10]

I don't know this 1 I'm sorry I can't help you

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3 years ago

"the toy car is about 3 inches long"is an example of a ________ observation

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It's a quantitative observation because it includes numerical data.

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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.

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

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3 years ago

A 0.05-kg car starts from rest at a height of 0.95 m. Assuming no friction, what is the kinetic energy of the car when it reache

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The kinetic energy of the car is A.) 0.466 J

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2 years ago

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