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neonofarm [45]
3 years ago
6

Two point charges, with charge magnitudes q and ????, are placed a distance r apart. In this arrangement, each point charge expe

riences a force of magnitude F. In the three scenarios, q, ????, and r are changed to some multiple of their original values. In each case, fill in the missing value on one of the variables that will result in a force between the point charges that is equal in magnitude to the original F.
Original q Q r
Scenario 1 2q ...Q 4r
Scenario 2 ...q 2Q (1/2)r
Scenario 3 (3/2)q (3/8)Q ...r
Physics
1 answer:
sammy [17]3 years ago
6 0

Answer:

1)  Q ’= 8 Q ,  2)    q ’= 16 q ,  3)   r ’= ¾ r

Explanation:

For this exercise we will use Coulomb's law

      F = k q Q / r²

It asks us to calculate the change of any of the parameters so that the force is always F

Original values

                q, Q, r

Scenario 1

      q ’= 2q

       r ’= 4r

     F = k q ’Q’ / r’²

we substitute

     F = k 2q Q ’/ (4r)²

     F = k 2q Q '/ 16r²

we substitute the value of F

      k q Q / r² = k q Q '/ 8r²

       Q ’= 8 Q

Scenario 2

       Q ’= Q

       r ’= 4r

we substitute

      F = k q ’Q / 16r²

      k q Q / r² = k q’ Q / 16 r²

      q ’= 16 q

Scenario 3

      q ’= 3/2 q

      Q ’= ⅜ Q

we substitute

        k q Q r² = k (3/2 q) (⅜ Q) / r’²

        r’² = 9/16 r²

        r ’= ¾ r

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Answer:

0.06 Kg

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Final velocity (v) = 3.0 m/s

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Net Force (F) = 3 N

Mass (m) =?

Next, we shall determine the acceleration of the object. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 3.0 m/s

Distance (s) = 0.09 m

Acceleration (a) =?

v² = u² + 2as

3² = 0² + (2 × a × 0.09)

9 = 0 + 0.18a

9 = 0.18a

Divide both side by 0.18

a = 9 / 0.18

a = 50 m/s²

Finally, we shall determine the mass of the object. This can be obtained as follow:

Net Force (F) = 3 N

Acceleration (a) = 50 N

Mass (m) =?

F = ma

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Therefore, the mass of the object is 0.06 Kg

5 0
2 years ago
An old millstone, used for grinding grain in a gristmill, is a solid cylindrical wheel that can rotate about its central axle wi
MAXImum [283]

Answer:

The answer is I=70,513kgm^2

Explanation:

Here we will use the rotational mechanics equation T=Ia, where T is the Torque, I is the Moment of Inertia and a is the angular acceleration.

When we speak about Torque it´s basically a Tangencial Force applied over a cylindrical or circular edge. It causes a rotation. In this case, we will have that T=Ft*r, where Ft is the Tangencial Forge and r is the radius

Now we will find the Moment of Inertia this way:

Ft*r=I*a -> (Ft*r)/(a) = I

Replacing we get that I is:

I=(200N*0,33m)/(0,936rad/s^2)

Then I=70,513kgm^2

In case you need to find extra information, keep in mind the Moment of Inertia for a solid cylindrical wheel is:  

I=(1/2)*(m*r^2)

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3 years ago
What is the maximum speed when the conditions are mass =450 kg, initial height= 30 m, and the roller coaster is initially at res
Zarrin [17]

Answer:

B. 24.2 m/s

Explanation:

Given;

mass of the roller coaster, m = 450 kg

height of the roller coaster, h = 30 m

The maximum potential energy of the roller coaster  due to its height is given by;

P.E_{max} = mgh\\\\PE_{max} = 450 *9.8*30\\\\PE_{max} = 132,300 \ J

P.E_{max} = K.E_{max} \ (law \ of \ conservation\ of \ energy)

K.E_{max} = \frac{1}{2}mv_{max}^2\\\\ v_{max}^2 = \frac{2K.E_{max}}{m}\\\\ v_{max}^2 = \frac{2*132300}{450}\\\\ v_{max}^2 =588\\\\v_{max} = \sqrt{588}\\\\  v_{max} = 24.2 \ m/s

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3 0
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1) The smallest part of an element that behaves like the element is the atom.
Mandarinka [93]

Here are the answers: 

1. False - Molecules is the smallest part of an element that behaves like the element. 

2. False - The nucleus contains both protons and neutrons

3. True

4. True

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6. D. Neutron

7. B. Protons and Neutrons

8. C. Electron

9. C. 6

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Marina CMI [18]

Explanation:

The equation of motion of an object is given by :

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We need to find the time when the object hits the ground. When the object hits the ground, h(t) = 0

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-16t^2+112t+128=0

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On solving above equation using online calculator, t = 8 seconds. So, the object hit the ground after 8 seconds. Hence, this is the required solution.

8 0
3 years ago
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