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3 years ago

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Physics

1 answer:

const2013 [10]3 years ago

5 0

Answer:

The Forces of Flight

At any given time, there are four forces acting upon an aircraft.

These forces are lift, weight (or gravity), drag and thrust. Lift is

the key aerodynamic force that keeps objects in the air. It is the

force that opposes weight; thus, lift helps to keep an aircraft in

the air. Weight is the force that works vertically by pulling all

objects, including aircraft, toward the center of the Earth. In order

to fly an aircraft, something (lift) needs to press it in the opposite

direction of gravity. The weight of an object controls how strong

the pressure (lift) will need to be. Lift is that pressure. Drag is a

mechanical force generated by the interaction and contract of a

solid body, such as an airplane, with a fluid (liquid or gas). Finally,

the thrust is the force that is generated by the engines of an

aircraft in order for the aircraft to move forward.

Explanation:

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gregori [183]

Answer:

a) $v=6.898\times 10^{6}\ m.s^{-1}$ is the speed of each proton

b) $F_c=3.97\times 10^{-13}\ N$

Explanation:

Given:

radius of path of motion, $r=0.2\ m$

we know charge on protons, $q=1.6\times 10^{-19}\ C$

magnetic field strength, $B=0.36\ T$

we've mass of proton, $m=1.67\times 10^{-27}\ kg$

From the equivalence of magnetic force and the centripetal force on the proton:

$F_B=F_C$

$q.v.B=\frac{m.v^2}{r}$

$q.B=\frac{m.v}{r}$

where:

v = speed of the proton

$(1.6\times 10^{-19})\times 0.36=\frac{1.67\times 10^{-27}\times v}{0.2}$

$v=6.898\times 10^{6}\ m.s^{-1}$ is the speed of each proton

Now the centripetal force on each proton:

$F_c=m.\frac{v^2}{r}$

$F_c=1.67\times 10^{-27}\times \frac{(6.898\times 10^6)^2}{0.2}$

$F_c=3.97\times 10^{-13}\ N$

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3 years ago

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3 years ago

A pulley with a radius of 3.0 cm and a rotational inertia of 4.5 x 10–3 kg∙m2 is suspended from the ceiling. A rope passes over

vekshin1

Answer:

22J

Explanation:

Given :

radius 'r'= 3cm

rotational inertia 'I'=4.5 x $10^{-3}$ kgm²

mass on one side of rope ' $m_{1$ '= 2kg

mass on other side of rope' $m_{2$ ' =4kg

velocity'v' of mass $m_{2$ ' = 2m/s

Angular velocity of the pulley is given by

‎ω = v /r => 2/ 3x $10^{-2$

‎ω = 66.67 rad/s

For the rotating body, we have

KE = $\frac{1}{2}$ I ω²

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$KE_{b} = \frac{1}{2} (m_1 + m_2).v^2\\KE_b= \frac{1}{2} (2+4).2^2$

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3 years ago

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Trava [24]

Answer:

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<h3>Define Induction?</h3>

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