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denis-greek [22]
3 years ago
9

A 4 kg block has a coefficient of friction of 0.3 between itself and the surface

Physics
1 answer:
Deffense [45]3 years ago
4 0

Answer:

C - 11.76 N (Newtons)

Explanation:

2 steps are needed,

First find the Force being applied by the box on the surface.

F = M(a) [Force equals: mass X Acceleration (of gravity)]

You get 39.2 N of Force.

Plug in the remaining values into the equation for Frictional Force

F(f) = μ(F) [Coefficient of Friction X Force applied by the Box]

So

F(f) = 0.3(39.2) = 11.76 N (of Frictional Force)

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Explanation:

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Consider the following mass distribution where the x- and y-coordinates are given in meters: 5.0 kg at (0.0, 0.0) m, 2.9 kg at (
Sauron [17]

Answer:

x = -1.20 m

y = -1.12 m

Explanation:

as we know that four masses and their position is given as

5.0 kg (0, 0)

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4 kg (2.5, 0)

8.3 kg (x, y)

As we know that the formula of center of gravity is given as

x_{cm} = \frac{m_1 x_1 + m_2x_2 + m_3x_3 + m_4x_4}{m_1 + m_2 + m_3 + m_4}

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10 + 8.3 x = 0

x = -1.20 m

Similarly for y direction we have

y_{cm} = \frac{m_1 y_1 + m_2y_2 + m_3y_3 + m_4y_4}{m_1 + m_2 + m_3 + m_4}

0 = \frac{5(0) + 2.9(3.2) + 4(0) + 8.3 y}{5 + 2.9 + 4 + 8.3}

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5 0
3 years ago
Please answer this. Science 7th grade.
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Answer:

the answer would be 2

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3 years ago
If two charged objects each have 2.5 C of charge on them and are located 100 m apart, how strong is the electrostatic force betw
LiRa [457]

Answer:

5.619×10⁶ N

Explanation:

Applying,

F = kqq'/r²................... Equation 1

Where F = electrostatic force between the charges, k = coulomb's constant, q = first charge, q' = second charge, r = distance btween the charges

From the questiion,

Given: q = 2.5 C, q' = 2.5 C, r = 100 m

Constant: 8.99×10⁹ Nm²/C²

Substitute these values into equation 1

F = (2.5×2.5×8.99×10⁹)/100²

F = 56.19×10⁵

F = 5.619×10⁶ N

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