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s344n2d4d5 [400]

2 years ago

8

The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.10. The mass of the object i

s 8.0 kg. What is the force of friction? What is the acceleration of the object?

1 answer:

denis-greek [22]2 years ago

6 0

Answer:

Explanation:

1) Force Friction = Normal Force * Coefficient of Friction

Force Friction = Mass * Gravity * Coefficient of Friction

2) F = ma

Force = mass * acceleration

Force Friction (from #1) = mass * acceleration

acceleration = Force Friction / Mass

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The four wheels of a car are connected to the car's body by spring assemblies that let the wheels move up and down over bumps an

maksim [4K]

Answer:

The approximate spring constant is $k = 55533.33 \ N/m$

Explanation:

From the question we are told that

The mass of the person is $m = 68 \ kg$

The dip of the car is $x = 1.2 \ cm = 0.012 \ m$

Generally according to hooks law

$F = k * x$

here the force F is the weight of the person which is mathematically represented as

$F = m * g$

=> $m * g = k * x$

=> $k = \frac{m * g }{x }$

=> $k = \frac{68 * 9.8}{ 0.012}$

=> $k = 55533.33 \ N/m$

8 0

3 years ago

An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m

r-ruslan [8.4K]

PART A)

Electrostatic potential at the position of origin is given by

$V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}$

here we have

$q_1 = 1.6 \times 10^{-19} C$

$q_2 = -1.6 \times 10^{-19} C$

$r_1 = r_2 = 1 m$

now we have

$V = \frac{Ke}{r} - \frac{Ke}{r}$

$V = 0$

Now work done to move another charge from infinite to origin is given by

$W = q(V_f - V_i)$

here we will have

$W = e(0 - 0) = 0$

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

$U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}$

$U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}$

$U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})$

$U = 1.15\times 10^{-30}$

Now we know

$KE = \frac{1}{2}mv^2$

$KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2$

$KE = 4.55 \times 10^{-27} kg$

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

4 0

2 years ago

A 5.0 kg block hangs from the ceiling by a mass-less rope. A Second block with a mass of 10.0 kg is attached to the first block

gayaneshka [121]

The tension in the first and second rope are; 147 Newton and 98 Newton respectively.

Given the data in the question

Mass of first block; $m_1 = 5.0kg$

Mass of second block, $m_2 =10kg$

Tension on first rope; $T_1 =\ ?$

Tension on second rope; $T_2 =\ ?$

To find the Tension in each of the ropes, we make use of the equation from Newton's Second Laws of Motion:

$F = m\ *\ a$

Where F is the force, m is the mass of the object and a is the acceleration ( In this case the block is under gravity. Hence ''a" becomes acceleration due to gravity $g = 9.8m/s^2$ )

For the First Rope

Total mass hanging on it; $m_T = m_1 + m_2 = 5.0kg + 10.0kg = 15.0kg$

So Tension of the rope;

$F = m\ * \ g\\\\F = 15.0kg \ * 9.8m/s^2\\\\F = 147 kg.m/s^2\\\\F = 147N$

Therefore, the tension in the first rope is 147 Newton

For the Second Rope

Since only the block of mass 10kg is hang from the second, the tension in the second rope will be;

$F = m\ * \ g\\\\F = 10.0kg \ * 9.8m/s^2\\\\F = 98 kg.m/s^2\\\\F = 98N$

Therefore, the tension in the second rope is 98 Newton

Learn More, brainly.com/question/18288215

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The wavelength of a wave on a string is 1.2 meters. If the speed of the wave is 60 meters/second, what is its frequency?

maria [59]

F=v/wavelength f=1.2/60=50Hz.

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2 years ago

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Est ce qu’un cricket c’est un consommateur

Sholpan [36]

Answer:

Oui

Explanation:

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