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olchik [2.2K]
3 years ago
12

A 117 kg horizontal platform is a uniform disk of radius 1.61 m and can rotate about the vertical axis through its center. A 62.

5 kg person stands on the platform at a distance of 1.05 m from the center, and a 28.3 kg dog sits on the platform near the person 1.43 m from the center. Find the moment of inertia of this system, consisting of the platform and its population, with respect to the axis.
Physics
1 answer:
Ivenika [448]3 years ago
8 0

Answer:

I_syst = 278.41477 kg.m²

Explanation:

Mass of platform; m1 = 117 kg

Radius; r = 1.61 m

Moment of inertia here is;

I1 = m1•r²/2

I1 = 117 × 1.61²/2

I1 = 151.63785 kg.m²

Mass of person; m2 = 62.5 kg

Distance of person from centre; r = 1.05 m

Moment of inertia here is;

I2 = m2•r²

I2 = 62.5 × 1.05²

I2 = 68.90625 kg.m²

Mass of dog; m3 = 28.3 kg

Distance of Dog from centre; r = 1.43 m

I3 = 28.3 × 1.43²

I3 = 57.87067 kg.m²

Thus,moment of inertia of the system;

I_syst = I1 + I2 + I3

I_syst = 151.63785 + 68.90625 + 57.87067

I_syst = 278.41477 kg.m²

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When you go out into the sun, the UV light can give you a tan. UV light has a frequency of 7.89 x 1014 Hz. What is the wavelengt
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Answer:

Explanation:

Formula and givens

  • λ = c / f
  • λ is the wavelength
  • c = the speed of light
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λ = ?

Solution

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What mass of electrons would be required to just neutralize the charge of 4.8 g of protons?
suter [353]
Let's calculate the total charge of M=4.8 g=0.0048 kg of protons.
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N_p =  \frac{M}{m_p}= \frac{0.0048 kg}{1.67 \cdot 10^{-27}kg}=2.87 \cdot 10^{24} And so the total charge of these protons is Q_p = qN_p = (1.6 \cdot 10^{-19}C)(2.87 \cdot 10^{24})=4.6\cdot 10^5 C

So, the neutralize this charge, we must have N_e electrons such that their total charge is
Q_e = -4.6 \cdot 10^5 C
Since the charge of each electron is q_e = -1.6 \cdot 10^{-19}C, the number of electrons needed is
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which is the same as the number of protons (because proton and electron have same charge magnitude). Since the mass of a single electron is m_e=9.1 \cdot 10^{-31}kg, the total mass of electrons should be
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6 0
3 years ago
A car moves with constant velocity along a straight road. Its position is x1 = 0 m at t1 = 0 s and is x2 = 66 m at t2 = 6.0 s .
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2. (9 points) A car starts from 10 mph and accelerates along a level road, i.e., no grade change. At 500 ft from its starting po
Sonja [21]

Answer:

a) t = 11.2 s

b) v = 70.5 mph

Explanation:

a)

  • Since we need to find the time, we could use the definition of acceleration (rearranging terms) as follows:

       t = \frac{v_{f} - v_{o}}{a}  (1)

  • where vf = 50 mph, and v₀ = 10 mph.
  • However, we still lack the value of a.
  • Assuming that the acceleration is constant, we can use the following kinematic equation:

       v_{f} ^{2} - v_{o} ^{2} = 2*a* \Delta x  (2)

  • Since we know that Δx = 500 ft, we could solve (2) for a.
  • In order to simplify things, let's first to convert v₀ and vf from mph to m/s, as follows:

       v_{o} = 10 mph*\frac{1609m}{1mi} *\frac{1h}{3600s} = 4.5 m/s  (3)

       v_{f} = 50 mph*\frac{1609m}{1mi} *\frac{1h}{3600s} = 22.5 m/s  (4)

  • We can do the same process with Δx, from ft to m, as follows:

       \Delta x = 500 ft *\frac{0.3048m}{1ft} = 152.4 m  (5)

  • Replacing (3), (4), and (5) in (2) and solving for a, we get:

       a = \frac{v_{f} ^{2} - v_{o}^{2}}{2*\Delta x} =  \frac{(22.5m/s) ^{2} - (4.5m/s)^{2}}{2*152.4m}  = 1.6 m/s2  (6)

  • Replacing (6) in (1) we finally get the value of the time t:

        t = \frac{v_{f} - v_{o}}{a} =  \frac{(22.5m/s) - (4.5m/s)}{1,6m/s2}  = 11.2 s  (7)

b)

  • Since the acceleration is constant, as we know the displacement is another 500 ft (152.4m), if we replace in (2) v₀ by the vf we got in a), we can find the new value of vf, as follows:

       v_{f} = \sqrt{v_{o} ^{2} +( 2*a* \Delta x)} = \sqrt{(22.5m/s)^{2} + (2*1.6m/s2*152.4m)} \\ v_{f} = 31.5 m/s (8)

  • If we convert vf again to mph, we have:

       v_{f} = 31.5m/s*\frac{1mi}{1609m} *\frac{3600s}{1h} = 70.5 mph  (9)

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