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Nataly [62]
2 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]2 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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Mechanical waves are those waves that  require a material medium for propagation.

<h3>What are mechanical waves?</h3>

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2 years ago
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3 years ago
The mass of the uniform cantilever is 1100 kg. Determine the force on the beam at A. Determine the force on the beam at B. Use C
Mashcka [7]

Answer:

Force A=-−2,697.75 N

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Explanation:

Taking moments at point A, the sum of clockwise and anticlockwise moments equal to zero.

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25*1100g=20Fb

Fb=25*1100g/20=1375g

Taking g as 9.81 then Fb=1375*9.81=13,488.75 N

The sum of upward and downward forces are same hence Fa=1100g-1375g=-275g

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Note that the centre of gravity is taken to be half the whole length hence half of 50 is 25 m because center of gravity is always at the middle

8 0
2 years ago
Calculate the change in length of concrete sidewalk (coefficient of linear expansion for concrete is 12*10^-6/celcius) that is 1
anyanavicka [17]

Answer:

The answer to your question is 5.4 cm

Explanation:

This problem refers to calculate the change in length in one dimension due to a change in temperature.

Data

α = 12 x 10⁻⁶

Lo = 150 meters

ΔT = 30 °C

Formula

                    ΔL/Lo = αΔT

solve for ΔL

                    ΔL = αLoΔT

Substitution

                    ΔL = (12 x 10⁻⁶)(150)(30)

Simplification

                    ΔL = 0054 m = 5.4 cm

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A car accelerates from 10m/s to 20m/s over a distance of 80m. What is its acceleration?
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Answer:

0.125m/s^2

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