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Artemon [7]
3 years ago
10

The 6kg box is pushed to the left at a constant speed. The coefficient of friction is 0.78. Solve for the amount of force with w

hich the hand pushes the box.
Physics
1 answer:
Dafna1 [17]3 years ago
4 0

Answer:

45.864N

Explanation:

Using the formula

F = nR

n is the coefficient of friction

R is the normal reaction

R = mg

F = nmg

F = 0.78 * 6 * 9.8

F = 45.864N

Hence the amount of force with which the hand pushes the box is 45.864N

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Each corner of a right-angled triangle is occupied by identical point charges "A", "B", and "C" respectively. Draw a sketch of t
NISA [10]

Answer:

Fnet = F√2

Fnet = kq²/r² √2

Explanation:

A exerts a force F on B, and C exerts an equal force F on B perpendicular to that.  The net force can be found with Pythagorean theorem:

Fnet = √(F² + F²)

Fnet = F√2

The force between two charges particles is:

F = k q₁ q₂ / r²

where

k is Coulomb's constant, q₁ and q₂ are the charges, and r is the distance between the charges.

If we say the charge of each particle is q, then:

F = kq²/r²

Substituting:

Fnet = kq²/r² √2

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3 years ago
An alpha particle (α), which is the same as a helium-4 nucleus, is momentarily at rest in a region of space occupied by an elec
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Answer:

Speed = 575 m/s

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Explanation:

Given :

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Mass of the alpha particle, $m_{\alpha} = 6.68 \times 10^{-27} \ kg$

Charge of the alpha particle is, $q_{\alpha} = 3.20 \times 10^{-19} \ C$

So the potential difference for the alpha particle when it is accelerated through the potential difference is

$U=\Delta Vq_{\alpha}$

And the kinetic energy gained by the alpha particle is

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From the law of conservation of energy, we get

$K.E. = U$

$\frac{1}{2}m_{\alpha}v_{\alpha}^2 = \Delta V q_{\alpha}$

$v_{\alpha} = \sqrt{\frac{2 \Delta V q_{\alpha}}{m_{\alpha}}}$

$v_{\alpha} = \sqrt{\frac{2(3.45 \times 10^{-3 })(3.2 \times 10^{-19})}{6.68 \times 10^{-27}}}$

$v_{\alpha} \approx 575 \ m/s$

The mechanical energy is conserved in the presence of the following conservative forces :

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5 0
3 years ago
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Answer:

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Explanation:

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borishaifa [10]

Answer:

8.854 pF

Explanation:

side of plate = 0.1 m ,

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V = 5 kV = 5000 V

V' = 1 kV = 1000 V

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