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Minchanka [31]
3 years ago
10

10. During 4th period we put Klaudia in a box because she was talking too much. We still heard her voice through the box so we d

ecided to push her outside. The force of friction of the ground on the box was 68 N. If Mr.Whitmore can apply a force of 25 N and every other 7th grade student can apply a force of 6 N. How many students would Mr. Whitmore need to make the box start moving and go outside. (Think quickly, the faster we move the box out, the quicker she stops talking)
Physics
1 answer:
Alex787 [66]3 years ago
6 0

Answer: Mr. Whitmore would need 7 or more students ( 7.17) to make the box start moving and go outside

Explanation:

Given that;

friction force of ground box = 68 N

student of 7th grade = n

Whitmore can apply a force of 25 N

every other 7th grade student can apply a force of 6 N.

now

friction force = forced applied by whitmore + total force ny 7th grade student

we substitute

68 = 25 + 6n

6n = 68 - 25

6n = 43

n = 43/6

n = 7.17

Therefore Mr. Whitmore would need 7 or more students ( 7.17) to make the box start moving and go outside

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4 years ago
Now let’s apply Coulomb’s law and the superposition principle to calculate the force on a point charge due to the presence of ot
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Answer:

F = - 1.68 10⁻⁴ N

, it is directed to the left of the x-axis

Explanation:

Coulomb's law is

     F = k q₁ q₂ / r²

Where K is the Coulomb constant that value 8.99 10⁹ N m²/ C², q are the electric charges and r is the distance between them. Let's apply to our problem for each pair of charges

Let's reduce the magnitudes to the SI system

    q₁ = 3.0 nc (1C / 10 9 nC) = 3.0 10⁻⁹ C

    x₁ = 2.0 cm (1m / 100cm) = 2.0 10⁻² m

    q₂ = -6.0 nC = -6.0 10⁻⁹ C

    x₂ = 4.0 cm = 4.0 10⁻² m

    q₃ = 5.0 nC = 5.0 10⁻⁹ C

    x3 = 0 m

Charges q1 and q3

    r = x₁ -x₃

    r = 2.0 10⁻² -0

    r = 2.0 10⁻² m

    F₁₃ = 8.99 10⁹ 3.0 10⁻⁹ 5.0 10⁻⁹ / (2.0 10⁻²)²

    F₁₃ = 33.7 10⁻⁵ N

As the charges are of the same sign, the force is repulsive, therefore it is directed to the left of the x-axis

Charges q2 and q3

    r = r₂ –r₃

    r = 4.0 10⁻² - 0 = 4.0 10⁻² m

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    F₂₃ = 16.86 10⁻⁵ N

As the charges are of different sign, the force is attractive, therefore it is directed to the right of the x-axis

The force is a vector magnitude, so each component must be added independently, in this case all the forces are on the x-axis, let's take the right direction as positive

    F = F₂₃ - F₁₃

    F = 16.86 10⁻⁵ - 33.7 10⁻⁵

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The negative sign means that it is directed to the left of the x-axis

5 0
4 years ago
The more diverse an ecosystem is —
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Answer:

C

Explanation:

5 0
3 years ago
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Two runners start at the same point on a straight track. The first runs with constant acceleration so that he covers 98 yards in
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Answer:

94.13 ft/s

Explanation:

<u>Given:</u>

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  • y = displacement to be moved by the rock during the time of flight along the vertical = 0 yard

<u>Assume:</u>

  • u = magnitude of initial velocity of the rock
  • \theta = angle of the initial velocity with the horizontal.

For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\

On dividing equation (1) by (2), we have

\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ

Now, putting this value in equation (2), we have

u\cos 51.34^\circ\times  5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s

Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.

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3 years ago
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3 years ago
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