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Sphinxa [80]
4 years ago
7

an airplanes propeller completes one revolution in 65ms. a)what is the angular speed in rpm? in rad/s?

Physics
1 answer:
miss Akunina [59]4 years ago
7 0

-- We already know the rate of revolutions per time ...
it's 1 revolution per 0.065 sec.  We just have to
unit-convert that to 'per minute'.

(1 rev / 0.065 sec) x (60 sec / min) = (1 x 60) / (0.065) = <em>923 RPM</em>  (rounded)
_______________________________

-- 1 revolution = 2π radians

  (2π rad) / (0.065 sec) = (2π / 0.065) = <em>96.66 rad/sec</em>  (rounded)


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An aircraft flying at a steady velocity of 70 m/s eastwards at a height of 800 m drops a package of supplies.
iren2701 [21]

The motion of the package can be described as the motion of a projectile,

given that it has an horizontal velocity and it is acted on by gravity.

  • a) \vec{v} = 70·i
  • b) The package will reach ground in approximately <u>12.77 seconds</u>.
  • c) The speed of the package as it lands is approximately <u>145.51 m/s</u>.
  • d) The path of the package based on a stationary frame of reference is <u>parabolic</u>
  • e) The path of the package as seen from the plane is <u>directly vertical</u> downwards

Reasons:

Velocity of the aircraft = 70 m/s

Direction of flight of the aircraft = Eastward

Height from which the aircraft drops the package, h = 800 m

a) The initial velocity of the package, \vec{v} = 70·i

b) The time it will take the package to reach the ground, <em>t</em>, is given by the formula;

\displaystyle h = \mathbf{\frac{1}{2} \cdot g \cdot t^2}

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore;

\displaystyle t = \mathbf{\sqrt{ \frac{2 \cdot h}{g} }}

Which gives;

\displaystyle t = \sqrt{ \frac{2 \times 800}{9.81} } \approx \mathbf{12.77}

The time it will take the package to reach the ground, t ≈ <u>12.77 seconds</u>

c) The vertical velocity just before the package reaches the ground, v_y, is given as follows;

v_y^2 = 2·g·h

Therefore;

v_y = √(2·g·h)

Which gives;

v_y = √(2 × 9.81 × 800) ≈ 125.28

v_y ≈ 125.28 m/s

Which gives; \vec{v} = 70·i - 125.28·j

Therefore, |v| = √(70² + (-125.28)²) ≈ 143.51

The speed of the package as it lands, |v| ≈ <u>143.51 m/s</u>

d) The motion of the package that includes both horizontal and vertical motion is a projectile motion.

Therefore;

The path of the package is the path of a projectile, which is a <u>parabolic shape</u>.

e) As seen by someone on the aeroplane, the horizontal velocity will be

zero, therefore, the package will appear as accelerating <u>directly vertical</u>

downwards.

Learn more about projectile motion here:

brainly.com/question/1130127

6 0
2 years ago
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Answer:

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Explanation:

Coulombs law states that "the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them."

Mathematically, F = kq1q2/r²

Where q1 and q2 are the charges

r is the distance between the charges.

According to the law, the force between two charged objects is related to the inverse of the square of the distance separating them.

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The answer is meters....I think??
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