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Masteriza [31]
3 years ago
6

What is the type of relationship between the height (H) and the atmospheric density, and which segment can give the most accurat

e approximations for the interpolated values?
(picture attached below)

Physics
1 answer:
vovikov84 [41]3 years ago
3 0

Answer:

B.  inverse plot, 0.51 kilograms/meter3

Explanation: First of all, we note that the relationship between the altitude and the atmospheric density is an inverse relationship. In fact, an inverse relationship is a relationship between the x-variable and the y-variable of the form

Therefore, as the x increases, the y decreases, and as the x decreases, they increases. This is exactly what occurs with the altitude and the atmospheric density in this plot: as the altitude increases, the density decreases, and vice-versa.

Moreover, we can infer the value of the atmospheric density at an altitude of 1,291 km. This point is located between point A (2550 km) and point B(1000 km), so the density must have a value between 0.30 kg/m^3 and 0.54 kg/m^3, so the correct choice is

B.  inverse plot, 0.51 kilograms/meter3

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b. Comparing and Contrasting Compare the change in atmospheric pressure with elevation to the change in water pressure with dept
olasank [31]

Answer:

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Explanation:

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4 0
3 years ago
Air resistance is an example of what type of friction?
Sav [38]
Well, Air resistance is a special type of friction (you cannot classify it in other categories). That force of air-resistance is often observed to oppose the motion of the object,( like every other frictional forces)

Hope this helps!
4 0
4 years ago
A person is trying to lift a crate that has a mass of 30 kg. The normal force of the floor is currently supplying 150N of force.
alexdok [17]

Here when an object is placed on the level floor then in that case there are two forces on the object

1). Weight of object downwards (mg)

2). Normal force due to floor which will counterbalance the weight (N)

so when no force is applied on the box at that time normal force is counter balanced by weight.

Now here it is given that A person tried to lift the box upwards

So now there are two forces on the box

1). Applied force of person

2). Normal force due to ground

So now these two forces will counter balance the weight of the crate

So we can write an equation for force balance like

F_g = F_n + F_a

given that

F_g = mg

here

m = 30 kg and

g = acceleration due to gravity = 10 m/s^2

F_n = 150 N

now from above equation

30*10 = 150 + F_a

F_a = 300 - 150 = 150 N

So force applied by the person must be 150 N

7 0
4 years ago
A 1.00-kg object is attached by a thread of negligible mass, which passes over a pulley of negligible mass, to a 2.00-kg object.
Anna [14]

Answer:

a = 3.27 m/s²

v = 2.56 m/s

Explanation:

given,

mass A = 1 kg

mass B = 2 kg

vertical distance between them = 1 m

F_d = mg

F_d = 2 \times 9.8

F_d = 19.6\ N

F_u = mg

F_u = 1 \times 9.8

F_u = 9.8\ N

F_{net} = 19.6 - 9.8

F_{net}=9.8\ N

F = (m_1+m_2)a

9.8 = (2+1)a

a = 3.27 m/s²

The speed of the system at that moment is:

v² = u² + 2×a×s

v² = 0² + 2× 3.27 × 1

v ² = 6.54

v = 2.56 m/s

3 0
3 years ago
According to Newton's 2nd Law of Motion, if the mass of an object is 10 kg and the force is 10 newtons, then the acceleration is
Mkey [24]

According to Newton's second Law of motion, if the mass of an object is 10 kg and the force is 10 newtons, then the acceleration is 1m/s².

<h3>How to calculate acceleration?</h3>

The acceleration of a moving body can be calculated by dividing the force of the body by its mass.

According to this question, the mass of an object is 10 kg and the force is 10 newtons, then the acceleration can be calculated as follows:

acceleration = 10N ÷ 10kg

acceleration = 1m/s²

Therefore, according to Newton's second Law of motion, if the mass of an object is 10 kg and the force is 10 newtons, then the acceleration is 1m/s².

Learn more about acceleration at: brainly.com/question/12550364

#SPJ1

8 0
2 years ago
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