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BARSIC [14]
2 years ago
5

A model of a helicopter rotor has four blades, each of length 3.0 m from the central shaft to the blade tip. The model is rotate

d in a wind tunnel at a rotational speed of 560 rev/min.part A :What is the linear speed of the bladetip?part B :What is the radial acceleration of the bladetip expressed as a multiple of the acceleration of gravity,g?
Physics
1 answer:
ohaa [14]2 years ago
4 0

Answer with Explanation:

We are given that

r=3 m

Angular frequency=\omega=560rev/min

A.1 revolution=2\pi radian

560 revolutions=560\times 2\pi rad

Angular frequency=2\times 3.14\times \frac{560}{60}rad/s

1 min=60 s

\pi=3.14

Angular frequency=\omega=58.6rad/s

Linear speed=\omega r

Using the formula

Linear speed=58.6\times 3=175.8m/s

Hence, the linear speed of the blade tip=175.8m/s

B.Radial acceleration=a_{rad}=\frac{v^2}{r}

By using the formula

Radial acceleration=a_{rad}=\frac{(175.8)^2}{3}= 10.301\times 10^3m/s^2

Radial acceleration=\frac{10.301}{9.8}g\times 10^3=1.05\times 10^3 g

Where g=9.8m/s^2

Hence, the radial acceleration=1.05\times 10^3 g

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<span>b) Easy = 1 meter </span>

<span>c) two choices 10m or 100 m . Go with 100 m </span>

<span>d) Stretch it out , trunk tip to tail tip - call it 10 m </span>

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8 0
3 years ago
Singly charged uranium-238 ions are accelerated through a potential difference of 2.90 kV and enter a uniform magnetic field of
Lera25 [3.4K]

Answer: 0.091 m

Explanation:

r = 1/B * √(2mV/e), where

r = radius of their circular path

B = magnitude of magnetic field = 1.29 T

m = mass of Uranium -238 ion = 238 * amu = 238 * 1.6*10^-27 kg

V = potential difference = 2.9 kV

e = charge of the Uranium -238 ion = 1.6*10^-19 C

r = 1/1.29 * √[(2 * 238 * 1.6*10^-27 * 2900) / 1.6*10^-19]

r = 1/1.29 * √(2.21*10^-21 / 1.6*10^-19)

r = 1/1.29 * √0.0138

r = 1/1.29 * 0.117

r = 0.091 m

Therefore, the radius of their circular path is 0.091 m

6 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS!!
Aneli [31]

d^2=4^2+3^2\\\\d^2=16+9\\\\d^2=25\\\\d=5

3 0
3 years ago
A car accelerates uniformly from rest to a speed of 22.4 km/h in 6.1 s. find the distance it travels during this time.
stepladder [879]
You will need to add 22.4+6.1
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3 years ago
A body from the top of a 180 m high mountain is abandoned. Disregarding air resistance and adopting g = 10m / s2, determine:
elena55 [62]

Answer:

1) 6 seconds

2) 60 m/s

Explanation:

Given:

Δy = 180 m

v₀ = 0 m/s

a = 10 m/s²

1) Find t.

Δy = v₀ t + ½ at²

180 m = (0 m/s) t + ½ (10 m/s²) t²

t = 6 s

2) Find v.

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (10 m/s²) (180 m)

v = 60 m/s

4 0
3 years ago
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