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4 years ago

If two objects with static charge are attracted what do you know about them?

Physics

1 answer:

klemol [59]4 years ago

4 0

They carry opposite charge ( one has negative charge and one has positive charge)

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Solve the inequality 2(n+3) – 4<6. Then graph<br> the solution.

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The solution is 22 2(n+3)-4&6

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3 years ago

What type of energy transformation occurs when a light bulb is turned on?

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A 1.0 kg copper rod rests on two horizontal rails 1.0 m apart and carries a current of 50 A from one rail to the other.

vagabundo [1.1K]

Answer

given,

mass of copper rod = 1 kg

horizontal rails = 1 m

Current (I) = 50 A

coefficient of static friction = 0.6

magnetic force acting on a current carrying wire is

F = B i L

Rod is not necessarily vertical

$F_x =i L B_d$

$F_y= i L B_w$

the normal reaction N = mg-F y

static friction f = μ_s (mg-F y )

horizontal acceleration is zero

$F_x-f = 0$

$iLBd = \mu_s(mg-F_y )$

B_w = B sinθ

B_d = B cosθ

iLB cosθ= μ_s (mg- iLB sinθ)

$B = \dfrac{\mu_smg}{i(cos\theta +\mu_s sin\theta)}$

$\theta =tan{-1}{\mu_s}$

$\theta =tan{-1}{0.6}$

$\theta = 31^0$

$B = \dfrac{0.6\times 1 \times 9.8}{50(cos31^0 +0.6 sin31^0)}$

B = 0.1 T

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3 years ago

A driver travels 225 miles in 4 hours and 30 minutes. What was his average speed in mph?

andrew-mc [135]

Answer:

195 is it current pls tell

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3 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':

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

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