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erica [24]
2 years ago
10

Glenn shoots an arrow at a 30.0 degree angle. It has a velocity of 65.0 m/s How far will the arrow travel?

Physics
1 answer:
Burka [1]2 years ago
3 0
I just shot my shot at this little 52 ms far
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Solid barium sulfate is placed into a beaker to form a saturated solution of barium sulfate. the solution has a barium concentra
borishaifa [10]
Ksp = [Ba⁺²][SO₄⁻²]

[Ba⁺²] = [SO₄⁻²] for barium sulfate
Thus,
Ksp = (1 x 10⁻⁵)²
Ksp = 1 x 10⁻¹⁰
5 0
2 years ago
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To practice Problem-Solving Strategy 23.2 for continuous charge distribution problems. A straight wire of length L has a positiv
Lesechka [4]

Answer:

             E = k Q / [d(d+L)]

Explanation:

As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field

       E = k ∫ dq/ r² r^

"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element  and "r^" is a unit ventor from the load element to the point.

Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant

         λ = Q / L

If we derive from the length we have

        λ = dq/dx       ⇒    dq = L dx

We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge

        dE = k dq / x²2

        dE = k λ dx / x²

Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider

        E = k \int\limits^{d+L}_d {\lambda/x^{2}} \, dx

We take out the constant magnitudes and perform the integral

        E = k λ (-1/x){(-1/x)}^{d+L} _{d}

   

Evaluating

        E = k λ [ 1/d  - 1/ (d+L)]

Using   λ = Q/L

        E = k Q/L [ 1/d  - 1/ (d+L)]

 

let's use a bit of arithmetic to simplify the expression

     [ 1/d  - 1/ (d+L)]   = L /[d(d+L)]

The final result is

     E = k Q / [d(d+L)]

3 0
3 years ago
A particle that carries a net charge of -41.8 μc is held in a region of constant, uniform electric field. the electric field vec
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A 3.30-kg block starts from rest at the top of a 30.0° incline and slides a distance of 2.10 m down the incline in 1.60 s.(a) fi
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