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frozen [14]
3 years ago
13

What statement is true about stable air?

Physics
2 answers:
Anettt [7]3 years ago
6 0

Answer: Option (4) is the correct answer.

Explanation:

A stable air is defined as the one which tends to resist movement. On the other hand, an unstable air will easily rise from one place to another.

By nature, stable air masses are calm and free of any violent disturbances. These are most of the time noticed by their appearance of stratiform clouds or fog.

Thus, we can conclude that the statement it tends to resist rising is true about stable air.

____ [38]3 years ago
4 0
The correct answer choice is 4
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One horsepower (hp) is the amount of power required to lift a 75-kg mass a vertical distance of 1 m in 1 s. What is 2 hp equival
Vladimir [108]

Answer:

1470 W

Explanation:

Power: This can be defined as the rate at which work is done or energy is used up. The S.I unit of power is Watt (W).

The expression for power is given as,

P = Energy/time

P = mgh/t ...................... Equation 1

Where P = power, m = mass, h = height, t = time, g = acceleration due to gravity.

Given: m = 75 kg, g =9.8 m/s², h = 1 m, t = 1 s.

Substitute into equation 1

P = (75×1×9.8)/1

P = 735 W.

From the above,

1 hp = 735 W

2 hp = (2×735) W

2 hp = 1470 W.

Hence 2 hp = 1470 W

8 0
3 years ago
Please help have due very soon?thank you
Andru [333]
The answer is :78 I think
8 0
2 years ago
ASAP! AWARDING BRAINLIEST see attached photo.
stellarik [79]

Answer:d

Explanation:

6 0
2 years ago
Read 2 more answers
A spring is used to stop a 50-kg package which is moving down a 20º incline. The spring has a constant k = 30 kN/m and is held b
Elina [12.6K]

Answer:

0.3 m

Explanation:

Initially, the package has both gravitational potential energy and kinetic energy.  The spring has elastic energy.  After the package is brought to rest, all the energy is stored in the spring.

Initial energy = final energy

mgh + ½ mv² + ½ kx₁² = ½ kx₂²

Given:

m = 50 kg

g = 9.8 m/s²

h = 8 sin 20º m

v = 2 m/s

k = 30000 N/m

x₁ = 0.05 m

(50)(9.8)(8 sin 20) + ½ (50)(2)² + ½ (30000)(0.05)² = ½ (30000)x₂²

x₂ ≈ 0.314 m

So the spring is compressed 0.314 m from it's natural length.  However, we're asked to find the additional deformation from the original 50mm.

x₂ − x₁

0.314 m − 0.05 m

0.264 m

Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.

8 0
2 years ago
A 50.0-kg bungee jumper steps off a bridge with a light bungee cord tied to her and to the bridge. The unstretched length of the
lina2011 [118]

Answer:

The the analysis for the free fall part should be done under the constant acceleration.

Explanation:

In the given problem, the jumper is falling under the free fall. Since, no external force is acting on the body therefore, the fall will be under the action gravity only. also, the acceleration due to gravity is always constant.

Therefore, the the analysis for the free fall part should be done under the constant acceleration.

8 0
2 years ago
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