Physics, 28.11.2019 01:31 leothedrifter

Afighter plane flying at constant speed 420 m/s and constant altitude 3300 m makes a turn of curvature radius 11000 m. on the ground, the plane’s pilot weighs (53 kg) (9.8 m/s 2 ) = 519.4 n. what is his/her apparent weight during the plane’s turn? answer in units of n

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Answer from: misrachel03

"Apparent weight during the "plan's turn" is 519.4 N

Explanation:

The "plane’s altitude" is not so important, but the fact that it is constant tells us that the plane moves in a "horizontal plane" and its "normal acceleration" is $\mathrm{a}_{\mathrm{n}}=\frac{v^{2}}{R}$

Given that,

v = 420 m/s

R = 11000 m

Substitute the values in the above equation,

$a_{n}=\frac{420^{2}}{11000}$

$a_{n}=\frac{176400}{11000}$

$a_{n}=16.03 \mathrm{m} / \mathrm{s}^{2}$

It has a horizontal direction. Furthermore, constant speed implies zero tangential acceleration, hence vector a = vector a N. The "apparent weight" of the pilot adds his "true weight" "m" "vector" "g" and the "inertial force""-m" vector a due to plane’s acceleration, vector $W_{\mathrm{app}}=m(\text { vector } g \text { -vector a })$

In magnitude,

$| \text { vector } g-\text { vector } a |=\sqrt{\left(g^{2}+a^{2}\right)}$

$| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=\sqrt{\left(9.8^{2}+16.03^{2}\right)}$

$| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=\sqrt{(96.04+256.96)}$

$| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=\sqrt{353}$

$| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=18.78 \mathrm{m} / \mathrm{s}^{2}$

Because vector “a” is horizontal while vector g is vertical. Consequently, the pilot’s apparent weight is vector

$\mathrm{W}_{\mathrm{app}}=(18.78 \mathrm{m} / \mathrm{s}^ 2)(53 \mathrm{kg})=995.77 \mathrm{N}$

Which is quite heavier than his/her true weigh of 519.4 N

Answer from: Quest

small = small current large mag large current

Answer from: Quest

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.

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