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Rainbow [258]
3 years ago
12

Two blocks connected by a rope of negligible mass are being dragged by a horizontal force (see figure below). Suppose F = 65.0 N

, m1 = 14.0 kg, m2 = 26.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.098
Determine the acceleration of the system

Physics
1 answer:
vovangra [49]3 years ago
7 0
Refer to the diagram shown below.

g = 9.8 m/s², and air resistance is ignored.

For mass m₁:
The normal reaction is m₁g.
The resisting force is R₁ = μm₁g.

For mass m₂:
The normal reaction is m₂g.
The resisting force is R₂ = μm₂g.

Let a =  the acceleration of the system.
Then
(m₁ + m₂)a = F - (R₁ + R₂)
(14+26 kg)*(a m/s²) = (65 N) - 0.098*(9.8 m/s²)*(14+26 kg)
40a = 65 - 38.416 = 26.584
a = 0.6646 m/s²

Answer:  0.665 m/s²  (nearest thousandth)

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8 0
1 year ago
Which of the following wavelengths will produce standing waves on a string that is 3.5 m long?
denpristay [2]

In a string of length L, the wavelength of the n-th harmonic of the standing wave produced in the string is given by:

\lambda=\frac{2}{n} L


The length of the string in this problem is L=3.5 m, therefore the wavelength of the 1st harmonic of the standing wave is:

\lambda=\frac{2}{1} \cdot 3.5 m=7.0 m


The wavelength of the 2nd harmonic is:

\lambda=\frac{2}{2} \cdot 3.5 m=3.5 m


The wavelength of the 4th harmonic is:

\lambda=\frac{2}{4} \cdot 3.5 m=1.75 m


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3 0
3 years ago
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Certain neutron stars (extremely dense stars) are believed to be rotating at about 500 rev/s. If such a star has a radius of 17
Alexus [3.1K]

Answer:

7.22 × 10²⁹ kg

Explanation:

For the material to be in place, the gravitational force on the material must equal the centripetal force on the material.

So, F = gravitational force = GMm/R² where M = mass of neutron star, m = mass of object and R = radius of neutron star = 17 km

The centripetal force F' = mRω² where R = radius of neutron star and ω  = angular speed of neutron star

So, since F = F'

GMm/R² = mRω²

GM = R³ω²

M = R³ω²/G

Since ω = 500 rev/s = 500 × 2π rad/s = 1000π rad/s = 3141.6 rad/s = 3.142 × 10³ rad/s and r = 17 km = 17 × 10³ m and G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²

Substituting the values of the variables into M, we have

M = R³ω²/G

M = (17 × 10³ m)³(3.142 × 10³ rad/s)²/6.67 × 10⁻¹¹ Nm²/kg²

M = 4913 × 10⁹ m³ × 9.872 × 10⁶ rad²/s²/6.67 × 10⁻¹¹ Nm²/kg²

M = 48,501.942 × 10¹⁵ m³rad²/s² ÷ 6.67 × 10⁻¹¹ Nm²/kg²

M = 7217.66 × 10²⁶ kg

M = 7.21766 × 10²⁹ kg

M ≅ 7.22 × 10²⁹ kg

8 0
2 years ago
Suppose that you are standing on a train accelerating at 0.20g. What minimum coefficient of static friction must exist between y
Ilia_Sergeevich [38]
Acceleration = (0.2 x g) = 1.96m/sec^2. 
<span>Accelerating force on 1kg. = (ma) = 1.96N. </span>
<span>1kg. has a weight (normal force) of 9.8N. </span>
<span>Coefficient µ = 1.96/9.8 = 0.2 minimum. </span>

<span>Coefficient is a ratio, so holds true for any value of mass to find accelerating force acting. </span>
<span>e.g. 75kg = (75 x g) = 735N. </span>
<span>Accelerating force = (735 x 0.2) = 147N</span>
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3 years ago
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