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Nataly [62]
3 years ago
15

An airplane is flying horizontally with a constant momentum during a time interval ?t. (a) Is there a net impulse acting on the

plane during this time? Use the impulse-momentum theorem to guide your thinking. (b) In the horizontal direction, both the thrust generated by the engines and air resistance act on the plane. Considering your answer to part (a), how is the impulse of the thrust related in magnitude and direction to the impulse of the force due to the air resistance?
Physics
1 answer:
goldenfox [79]3 years ago
7 0

Answer:

Explanation:

Given

Airplane is flying with horizontally with a constant momentum during time interval \Delta t

Impulse is given by change in momentum, so there is no net impulse on the Plane because momentum is constant

(b) As there is no change in momentum therefore impulse of thrust and air drag is balanced i.e. both are equal in magnitude but act in opposite direction                            

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On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the
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Answer:

a) FE = 0.764FG

b) a = 2.30 m/s^2

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a) To compare the gravitational and electric force over the particle you calculate the following ratio:

\frac{F_E}{F_G}=\frac{qE}{mg}              (1)

FE: electric force

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g: gravitational acceleration = 9.8 m/s^2

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m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg

You replace the values of all parameters in the equation (1):

\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764

Then, the gravitational force is 0.764 times the electric force on the particle

b)

The acceleration of the particle is obtained by using the second Newton law:

F_E-F_G=ma\\\\a=\frac{qE-mg}{m}

you replace the values of all variables:

a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

7 0
3 years ago
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now from above formula we have

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What is the frequency of a wave that has a speed of 5 m/s and a wavelength of 0.5meter?
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Explanation:

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