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nirvana33 [79]
3 years ago
11

An aircraft flying from a region of warmer air towards a region of colder air will _____ altitude, and the altimeter will read a

n altitude _____than the plane’s true elevation, unless corrections are made to the altimeter.
Physics
2 answers:
Step2247 [10]3 years ago
5 0

Answer:

lose, higher

Explanation:

An aircraft lose altitude when flying from a region of higher temperature to a region of cooler temperature, hence making the reading on the altimeter to be high.

olga2289 [7]3 years ago
4 0

Answer:

An aircraft flying from a region of warmer air towards a region of colder air will _____ altitude, and the altimeter will read an altitude _____than the plane’s true elevation, unless corrections are made to the altimeter.

A. lose; lower

B. gain; lower

C. lose; higher

D. gain; higher

The correct option is 'B' gain; lower

Explanation:

When an aircraft flies from a region of warm air to a region of colder air there is an increase in the altitude,  caused by the pressure created at these elevations. The altimeter will not show this increase in altitude and will remain constant, thus the reading of altimeter is lower when compared to the true elevation of the plane.

With the process happening repeatedly when the aircraft flies from the region of warm to cold air, the altimeter will read a higher value.  On the other hand, if the aircraft flies from a region of cold air to warm air, the altimeter will read a lower value.

The altimeter reading can be made accurate if corrected.

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A horizontal force of 12N is applied to 1.5kg of block which rests on a horizontal surface. If the coefficient friction is 0.3,
Rama09 [41]

Answer:

7 m/s²

Explanation:

From the question given above, the following data were:

Force applied (Fₐ) = 15 N

Mass (m) of block = 1.5 Kg

Acceleration due to gravity (g) = 10 m/s²

Coefficient of friction (μ) = 0.3

Acceleration (a) of block =?

Next, we shall determine the frictional force. This can be obtained as follow:

Mass (m) of block = 1.5 Kg

Acceleration due to gravity (g) = 10 m/s²

Coefficient of friction (μ) = 0.3

Normal reaction (R) = mg = 1.5 × 10 = 15 N

Frictional force (Fբ) =?

Fբ = μR

Fբ = 0.3 × 15

Fբ = 4.5 N

Next, we shall determine the net force acting on the block. This can be obtained as follow:

Force applied (Fₐ) = 15 N

Frictional force (Fբ) = 4.5 N

Net force (Fₙ) =.?

Fₙ = Fₐ – Fբ

Fₙ = 15 – 4.5

Fₙ = 10.5 N

Finally, we shall determine the acceleration produced in the block. This can be obtained as follow:

Mass (m) of block = 1.5 Kg

Net force (Fₙ) = 10.5 N

Acceleration (a) of block =?

Fₙ = ma

10.5 = 1.5 × a

Divide both side by a

a = 10.5 / 1.5

a = 7 m/s²

Therefore, the acceleration produced in the block is 7 m/s²

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