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Nikolay [14]
3 years ago
11

A 10.0-kg box starts at rest on a level floor. An external, horizontal force of 2.00 × 102 N is applied to the box for a distanc

e of 4.00 m. If the coefficient of kinetic friction between the box and the floor is 0.440, how fast is the block moving at the end of the 4.00 m? Assume it starts from rest.
Physics
1 answer:
Harman [31]3 years ago
7 0

Answer:

vf = 11.2 m/s

Explanation:

m = 10 Kg

F = 2*10² N

x = 4.00 m

μ = 0.44

vi = 0 m/s

vf = ?

We can apply Newton's 2nd Law

∑ Fx = m*a   (→)

F - Ffriction = m*a  ⇒  F - (μ*N) = F - (μ*m*g) = m*a   ⇒  a = (F - μ*m*g)/m

⇒    a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²

then , we use the equation

vf² = vi² + 2*a*x    ⇒    vf = √(vi² + 2*a*x)

⇒   vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s

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Answer:

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Explanation:

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And the important part is that <u>they do not touch one another</u>, but get close to each other.

In this case, the end of the neutral rod which is the closest part to the charged rod would acquire a negative charge. This is because of the Coulomb's Law. The opposite charges exert an attractive force to each other. The positive charges attract the negative charges on the neutral rod.

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A thin rod of length 0.75 m and mass 0.42 kg is suspended freely from one end. It is pulled to one side and then allowed to swin
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Answer:

(A) 0.63 J  

(B) 0.15 m

Explanation:

length (L) = 0.75 m

mass (m) =0.42 kg

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To solve the questions (a) and (b) we first need to calculate the rotational inertia of the rod (I)

I = Ic + mh^{2}  

Ic is the rotational inertia of the rod about an axis passing trough its centre of mass and parallel to the rotational axis

h is the horizontal distance between the center of mass and the rotational axis of the rod

I = (\frac{1}{12})(mL^{2} ) + m([tex]\frac{L}{2})^{2}[/tex]

I = (\frac{1}{12})(0.42 x 0.75^{2} ) + ( 0.42 x ([tex]\frac{0.75}{2})^{2}[/tex])

I = 0.07875 kg.m^{2}

(A) rods kinetic energy = 0.5Iω^{2}

  = 0.5 x 0.07875 x 4^{2} = 0.63 J   0.15 m

(B) from the conservation of energy

   initial kinetic energy + initial potential energy = final kinetic energy + final potential energy

   Ki + Ui = Kf + Uf

   at the maximum height velocity = 0 therefore final kinetic energy = 0

   Ki + Ui = Uf

   Ki = Uf - Ui

 Ki =  mg(H-h)

where (H-h) = rise in the center of mass

     0.63 = 0.42 x 9.8 x (H-h)

   (H-h) = 0.15 m

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What and where is the asteroid belt?<br><br> Please ANSWER THIS
Rudiy27

Answer:

The asteroid belt is a region of our solar system, between the orbits of Mars and Jupiter, in which many small bodies orbit our sun.

Explanation:

Hope this helps!

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Answer:

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Explanation:

The impedance of an LCR circuit shown as

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Variation of Z with respect to υ is shown in the figure.

As υ increases, Z decreases and so the current increases.

At υ = υ↓r

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​Z increases and so current decreases.

so the combination of circuit elements that is most suitable to comprise

the circuit is R, L and C.

To learn more about these circuits

brainly.com/question/13140756

#SPJ4

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