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2 years ago

10

An aircraft performs a maneuver called an "aileron roll." During this maneuver, the plane turns like a screw as it maintains a s

traight flight path, which sets the wings in circular motion. If it takes it 35 s to complete the circle and the wingspan of the plane is 11 m, what is the acceleration of the wing tip?

1 answer:

Dmitriy789 [7]2 years ago

3 0

Answer:

0.17724 m/s²

Explanation:

D = Diameter of roll = Length of wing = 11 m

T = Time it takes to complete the circle = 35 s

Velocity

$v=\frac{2\pi R}{T}\\\Rightarrow v=\frac{\pi D}{T}\\\Rightarrow v=\frac{\pi\times 11}{35}\\\Rightarrow v=0.98735\ m/s$

Acceleration

$a=\frac{v^2}{R}\\\Rightarrow a=\frac{0.98735^2}{\frac{11}{2}}\\\Rightarrow a=0.17724\ m/s^2$

Acceleration of the tip of the plane is 0.17724 m/s²

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A strong lightning bolt transfers an electric charge of about 31 C to Earth (or vice versa). How many electrons are transferred?

olchik [2.2K]

Answer:

$m=5.78\times 10^{-3}\ g$

$n_e=1.935\times 10^{20}$ is the no. of electrons

Explanation:

Given:

quantity of charge transferred, $Q=31\ C$

<u>No. of electrons in the given amount of charge:</u>

As we have charge on one electron $1.602\times 10^{-19}\ C$

so,

$n_e=\frac{Q}{e}$

$n_e=\frac{31}{1.602\times 10^{-19}}$

$n_e=1.935\times 10^{20}$ is the no. of electrons

Now if each water molecules donates one electron:

Then we require $n=1.935\times 10^{20}$ molecules.

<u>Now the no. of moles in this many molecules:</u>

$n_m=\frac{n}{N_A}$

where

$N_A=6.022\times 10^{23}$ Avogadro No.

$n_m=\frac{1.935\times 10^{20}}{6.022\times 10^{23}}$

$n_m=3.213\times 10^{-4}\ moles$

We have molecular mass of water as M=18 g/mol.

<u>So, the mass of water in the obtained moles:</u>

$n_m=\frac{m}{M}$

where:

m = mass in gram

$3.213\times 10^{-4}=\frac{m}{18}$

$m=5.78\times 10^{-3}\ g$

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2 years ago

A point charge q1 = 3.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 10 m.

UkoKoshka [18]

Answer:

a) 1.6 mN b) -1.6 mN c) -1.6 mN d) 1.6 mN

Explanation:

The electrostatic force between 2 point charges, obeys the Coulomb's Law, that can be expressed as follows:

F₁₂ = k*q₁*q₂/(r₁₂)² (in magnitude)

The direction of the force, is along the line that joins the charges (along the x axis) and as q₁ and q₂ are of the same sign, aims away from both charges.

a) So, for the force on q₂, we have:

F₁₂ = 9*18*10⁻⁵ N = 1.6 mN (positive as it is aiming in the positive x direction)

b) The force on q1, according to Newton's 3rd Law, is just equal and opposite to the one on q2:

F₂₁ = (-9*18*10⁻⁵) N = -1.6 mN (towards the negative x direction, away from q1)

c) If q₂ were -6.0 μC, the force will be the same in magnitude, but as now both charges have different signs, they wil attract each other, so the direction of the forces will be exactly the opposite to the first case:

F₁₂ = -1.6 mN (going towards the origin, where q₁ is located)

F₂₁ = 1.6 mN (going in the positive x direction, towards q₂)

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3 years ago

Read 2 more answers

A car is accelerating at 30 m/s2, if the car is 400 kg how much force

Verizon [17]

It would be 12,000 because newton’s third 2nd law states F=ma (force=matter x acceleration) so 30x400 would be your force .

please mark brainliest and i hope this helps!

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2 years ago

One of the key disadvantages of buying fitness equipment is __________. A. privacy B. flexibility C. maintenance D. choice Pleas

Alex Ar [27]

Answer:

<h2>C. maintenance </h2>

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A racquetball strikes a wall with a speed of 30 m/s and rebounds in the opposite direction with a speed of 26 m/s. The collision

Fudgin [204]

Answer:

The average acceleration of the ball during the collision with the wall is $a=2,800m/s^{2}$

Explanation:

<u>Known Data</u>

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<u>Initial momentum</u>

$p_{i}=mv_{i}=(-30m/s)(m_{b})=-30m_{b}\ m/s$

<u>final momentum</u>

$p_{f}=mv_{f}=(26m/s)(m_{b})=26m_{b}\ m/s$

<u>Impulse</u>

$I=\Delta p=p_{f}-p_{i}=26m_{b}\ m/s-(-30m_{b}\ m/s)=56m_{b}\ m/s$

<u>Average Force</u>

$F=\frac{\Delta p}{\Delta t} =\frac{56m_{b}\ m/s}{0.020s} =2800m_{b} \ m/s^{2}$

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$F=ma$ , so $a=\frac{F}{m_{b}}$ .

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3 years ago