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sladkih [1.3K]
3 years ago
10

The diameters of the main rotor and tail rotor of a single-engine helicopter are 7.56 m and 1.00 m, respectively. The respective

rotational speeds are 453 rev/min and 4137 rev/min. Calculate the speeds of the tips of both rotors.
Physics
1 answer:
valentina_108 [34]3 years ago
7 0

Answer:

Explanation:

Given

Diameter of main rotor d_1=7.56 m

Tail rotor d_2=1 m

N_1=453 rev/min

N_2=4137 rev/min

\omega =\frac{2\pi N}{60}

\omega _1=\frac{2\pi 453}{60}=47.44 rad/s

\omega _2=\frac{2\pi 4137}{60}=433.28 rad/s

Speed of the tip of main rotor=\omega _1\times r_1=47.44\times \frac{7.56}{2}=179.32 m/s

Speed of tail rotor=\omega _2\times r_2=433.28\times \frac{1}{2}=216.64 m/s

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A 1500 kg tractor pulls a 750 kg trailer north and applies a 2250 N force on it. What is the force on the tractor?
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Answer:

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Explanation:

The tractor and the trailer are two bodies that interact, therefore, by the law of Action and Reaction, the force that one applies on the other is equal to the force that the second body (trailer) applies on the first (tractor), but with opposite direction

          F = 2250 N

directed from trailer to tractor

3 0
3 years ago
Plz help asap.
stich3 [128]

Answer:

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Unlike a primary cell, a dry cell contains paste of an electrolyte instead of the solution. The contents of electrolyte paste react with each other through a chemical process and convert the chemical energy into electrical energy.

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A 1700 W laser emits light with a wavelength of 700 nm into a 3.0 mm diameter beam. What force does the laser beam exert on a co
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Answer:

F=5.7×10⁻⁶

Explanation:

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In a simplified model of a hydrogen atom, the electron moves around the proton nucleus in a circular orbit of radius 0.53×10−10m
Ksenya-84 [330]

Answer

given,

radius of the circular orbit, r = 0.53 x 10⁻¹⁰ m

mass of electron, M = 9.11 x 10⁻³¹ Kg

charge of electron, q₁ = 1.6 x 10⁻¹⁹ C

                                q₂ = 1.6 x 10⁻¹⁹ C

we know, force between two charges

F = \dfrac{kq_1q_2}{r^2}

F = \dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(0.53\times 10^{-10})^2}

  F = 8.20 x 10⁻⁸ N

b) using newton's second law

F = m a

m a =  8.20 x 10⁻⁸

a =\dfrac{8.20\times 10^{-8}}{9.11\times 10^{-31}}

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c) speed of the electron

 a =\dfrac{v^2}{r}

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d) the period of the circular motion.

    T=\dfrac{2\pi}{\omega}

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8 0
3 years ago
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If a positively charged ion is more concentrated outside the cell, the forces required to balance the chemical gradient would be
galben [10]

Answer: If a positively charged ion is more concentrated outside the cell, the forces required to balance the chemical gradient would be directed OUTWARD. Thus, the equilibrium potential for this ion would be POSITIVELY charged. The correct answer is OUTWARD: POSITIVELY.

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Therefore If a positively charged ion is more concentrated outside the cell, the forces required to balance the chemical gradient would be directed outward.Thus, the equilibrium potential for this ion would be positively charged.

8 0
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