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3 years ago

HELP!! What propeller design gives off the most air??

1 answer:

6 0

In aeronautics, a propeller, also called an airscrew, converts rotary motion from an engine or other power source into a swirling slipstream which pushes the propeller forwards or backwards. It comprises a rotating power-driven hub, to which are attached several radial airfoil-section blades such that the whole assembly rotates about a longitudinal axis. The blade pitch may be fixed, manually variable to a few set positions, or of the automatically variable "constant-speed" type.

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An asteroid is discovered in a nearly circular orbit around the Sun, with an orbital radius that is 3.570 times Earth's. What is

inysia [295]

Explanation:

Given:

$r_a = 3.570R_E$

$R_E = 1.499×10^{11}\:\text{m}$

$M_S = 1.989×10^{30}\:\text{kg}$

$G = 6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2$

Let $m_s$ = mass of the asteroid and $r_a$ = orbital radius of the asteroid around the sun. The centripetal force $F_c$ is equal to the gravitational force $F_G:$

$F_c = F_G \Rightarrow m_a\dfrac{v_a^2}{r_a} = G\dfrac{m_aM_S}{r_a^2}$

$\dfrac{4\pi^2 r_a}{T^2} = G\dfrac{M_S}{r_a^2}$

where

$v = \dfrac{2\pi r_a}{T}$

with T = period of orbit. Rearranging the variables, we get

$T^2 = \dfrac{4\pi^2 r_a^3}{GM_S}$

Taking the square root,

$T = 2\pi \sqrt{\dfrac{r_a^3}{GM_S}}$

$\:\:\:\:=2\pi \sqrt{\dfrac{(3.57(1.499×10^{11}\:\text{m}))^3}{(6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2)(1.989×10^{30}\:\text{kg})}}$

$\:\:\:\:= 2.13×10^8\:\text{s} = 6.75\:\text{years}$

3 0

3 years ago

What does the word deficiency mean a.need b.excess c.problem d.lack

Dahasolnce [82]

Answer is D! Ex. Andy has a Vitamin A deficiency.

6 0

3 years ago

The greater the mass of an object? I’ll give brainliest!*

liraira [26]

between B and C im thinking its more C though

6 0

3 years ago

A 0.1 kg toy contains a compressed spring. when the spring is released the toy fly 0.45 m upwards from ground level before falli

trapecia [35]

The speed of the toy when it hits the ground is 2.97 m/s.

The given parameters;

mass of the toy, m = 0.1 kg

the maximum height reached by the, h = 0.45 m

The speed of the toy before it hits the ground will be maximum. Apply the principle of conservation of mechanical energy to determine the maximum speed of the toy.

P.E = K.E

$mgh_{max} = \frac{1}{2} mv_{max}^2\\\\gh_{max} = \frac{1}{2} v_{max}^2\\\\v_{max}^2= 2gh_{max}\\\\v_{max} = \sqrt{2gh_{max}}$