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AlladinOne [14]
4 years ago
8

A 50-kg man is driving his snowmobile at a constant speed. The engine applies a force of 360 N and the coefficient of kinetic fr

iction between the snowmobile and the snow is 0.2. What is the mass of the snowmobile?
Physics
1 answer:
S_A_V [24]4 years ago
8 0

Answer:

134 kg

Explanation:

The snowmobile is moving at constant speed, so the acceleration is 0.

Sum of the forces in the x direction:

∑F = ma

F − 360 = 0

F = 360

Friction force equals the normal force times the coefficient of friction:

Nμ = 360

On level ground, the normal force equals the combined weight:

(m + 50)g μ = 360

m + 50 = 360 / (gμ)

m = -50 + 360 / (gμ)

m = -50 + 360 / (9.8 × 0.2)

m ≈ 134 kg

Round as needed.

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3 years ago
Eventually, the process approaches a steady state. In that steady state, the charge of the capacitor is not changing. What is th
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Answer:

The current in the circuit must be zero.

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In a RC circuit, the steady state is reached when either the capacitor is fully charged or fully discharged. In either case, there must not be any current through the circuit because if it exists, it will deliver charge to the capacitor and thus change its charge, which is not a steady state.

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3 years ago
Read 2 more answers
Arrange the examples in order, starting with the object that has the least amount of energy. In each case, assume there’s no fri
Artemon [7]
First example: book, m= 0.75 kg, h=1.5 m, g= 9.8 m/s², it has only potential energy Ep,

Ep=m*g*h=0.75*9.8*1.5=11.025 J

Second example: brick, m=2.5 kg, v=10 m/s, h=4 m, it has potential energy Ep and kinetic energy Ek,

E=Ep+Ek=m*g*h + (1/2)*m*v²=98 J + 125 J= 223 J

Third example: ball, m=0.25 kg, v= 10 m/s, it has only kinetic energy Ek

Ek=(1/2)*m*v²=12.5 J.

Fourth example: stone, m=0.7 kg, h=7 m, it has only potential energy Ep,

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The order of examples starting with the lowest energy:

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4 0
3 years ago
By calculating its wavelength (in nm), show that the second line in the Lyman series is UV radiation.
Rashid [163]

Answer:

 λ = 102.78  nm

This radiation is in the UV range,

Explanation:

Bohr's atomic model for the hydrogen atom states that the energy is

           E = - 13.606 / n²

where 13.606 eV   is the ground state energy and n is an integer

an atom transition is the jump of an electron from an initial state to a final state of lesser emergy

            ΔE = 13.606 (1 / n_{f}^{2} - 1 / n_{i}^{2})

the so-called Lyman series occurs when the final state nf = 1, so the second line occurs when ni = 3, let's calculate the energy of the emitted photon

            DE = 13.606 (1/1 - 1/3²)

            DE = 12.094 eV

let's reduce the energy to the SI system

            DE = 12.094 eV (1.6 10⁻¹⁹ J / 1 ev) = 10.35 10⁻¹⁹ J

let's find the wavelength is this energy, let's use Planck's equation to find the frequency

            E = h f

             f = E / h

            f = 19.35 10⁻¹⁹ / 6.63 10⁻³⁴

            f = 2.9186 10¹⁵ Hz

now we can look up the wavelength

           c = λ f

           λ = c / f

           λ = 3 10⁸ / 2.9186 10¹⁵

           λ = 1.0278  10⁻⁷ m

let's reduce to nm

            λ = 102.78  nm

This radiation is in the UV range, which occurs for wavelengths less than 400 nm.

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3 years ago
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