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algol [13]
3 years ago
10

QUESTION 1 67.2 kilometers = how many decameters

Physics
2 answers:
sweet-ann [11.9K]3 years ago
8 0

Answer:

67.2 kilometers =6720 decamaters

den301095 [7]3 years ago
6 0

Answer: the answer should be 6,720 decameters.

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Avogadro's number: 6.02 x 10^23 atoms is present in 1mol of a solid (i.e. 22, 400 cm3)

Hence, in 1 cm3, 6.02 x 10^23 /22400 atoms is present = 2 x 10 ^ 19 atoms.
7 0
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How long a star lives and what it becomes at the end of its life depends primarly on what
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4 0
3 years ago
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
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tekilochka [14]

The gravitational force between two object depends on their masses and on their distance.


Since the formula is


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If the masses grow, the force also grows. But I'm assuming the two objects are fixed, so you can't enlarge their mass.


So, the only option remaining is to lower their distance: since it sits at the denominator, a smaller value of d results in a bigger value for F.


So, if you reduce the distance between two objects, the gravitational force between them will always result in an increase

6 0
3 years ago
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