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Lera25 [3.4K]
3 years ago
8

A jet plane leaves the stratosphere and returns to the ground. How does the temperature outside the jet change during its descen

t through the troposphere?   A. It decreases and then increases.   B. It decreases steadily.   C. It increases steadily.   D. It increases and then decreases.

Physics
2 answers:
finlep [7]3 years ago
8 0

Answer : A. It decreases and then increases.

Explanation :  Troposphere is the lowermost layer of atmosphere.

Stratosphere is next layer up to the troposphere. As the jet descends from stratosphere towards the troposphere, the temperature initially decreases and then at troposphere is roughly constant and then steadily increases.

So, option (A) is correct.

inn [45]3 years ago
3 0

Answer:

The answer is not A It decreases and then increases. The correct answer is C It increases steadily.

Explanation: I just got the answer wrong by going with It decreases and then increases.

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A steel beam of mass 1975 kg and length 3 m is attached to the wall with a pin that can rotate freely on its right side. A cable
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Answer:

a) 29062.125 N·m

b) 0 N·m

c) Torque, due \ to \ tension =L\cdot Tsin\theta = \frac{M\cdot L\cdot g}{2}

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Explanation:

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Beam length = 3 m

Cable angle = 60° above horizontal

a) We have the formula for torque given as follows;

Torque about the pin = Force × Perpendicular distance of force from pin

Where the force = Force due to gravity or weight, we have

Weight = Mass × Acceleration due to gravity = 1975 × 9.81 = 19374.75 N

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Torque due to gravity = 19374.75 N × 1.5 m = 29062.125 N·m

b) Torque about the pinned end due to the contact forces between the pin and the beam is given by the following relation;

Since the distance from pin to the contact forces between the pin and the beam is 0, the torque which is force multiplied by perpendicular distance is also 0 N·m

c) To find the expression for the tension force, T we find the sum of the moment forces about the pin as follows

Sum of moments about p is given as follows

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T·sin(θ) × L= M×L/2×g

Therefore torque due to tension is given by the following expression

Torque, due \ to \ tension =L\cdot Tsin\theta = \frac{M\cdot L\cdot g}{2}

d) Plugging in the values in the torque due to tension equation, we have;

3\times Tsin60 = \frac{1975\times 3\times 9.81}{2} = 29062.125

Therefore, we make the tension force, T the subject of the formula hence

T= \frac{29062.125}{3 \times sin(60)} = 11186.02 N

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