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Reika [66]
2 years ago
13

Ricardo is on vacation, doing some mountain climbing. He notices that the higher he goes up a mountain, the colder he feels. He

remembers his physics teacher teaching about these types of relationships. What is the type of relationship between mountain elevation and temperature?
Physics
2 answers:
Irina-Kira [14]2 years ago
8 0

Answer:

B. an inverse relationship

Explanation:

Here is the complete question

Ricardo is on vacation, doing some mountain climbing. He notices that the higher he goes up a mountain, the colder he feels. He remembers his physics teacher teaching about these types of relationships. What is the type of relationship between mountain elevation and temperature? A. a positive relationship B. an inverse relationship C. a neutral relationship D. a direct relationship

Solution

It is an inverse relationship because, as Ricardo's mountain elevation increases, he feels colder. So, as his mountain elevation increases, the temperature decreases.

Since one variable decreases while the other increases, it can only be an inverse relationship.

Let h be Ricardo's mountain elevation and T his temperature. So by inverse proportionality,

h ∝ 1/T

h = k/T

hT = k = constant

So, we have an inverse relationship and B is the answer.

cupoosta [38]2 years ago
7 0

Answer:

An inverse relationship.

Explanation:

Just took the quiz on edgenuity.

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A 60 kg acrobat is in the middle of a 10 m long tightrope. The center of the rope dropped 30 cm in relation to the ends that are
Zigmanuir [339]

Answer:

The tension in each half of the rope, is approximately 4,908.8 N

Explanation:

The mass of the acrobat, m = 60 kg

The length of the rope, l = 10 m

The extent by which the center dropped = 30 cm = 0.3 m

Let, 'T' represent the tension in each half of the rope

Weight, W = Mass, m × The acceleration due to gravity, g

∴ W = m × g

The acceleration due to gravity, g ≈ 9.8 m/s²

∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N

The angle the dropped rope makes with the horizontal, θ is given as follows;

θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°

At equilibrium, the sum of vertical forces, \Sigma F_y = 0

The vertical component of the tension, T_y, in each half of the rope is given as follows;

T_y = T × sin(θ)

∴ \Sigma F_y = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)

Plugging in the values, with θ = arctan(0.06) for accuracy, we get;

588 N + 2 × T × sin(arctan(0.06) = 0

∴ 2 × -T × sin(arctan(0.06) = 588 N

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The tension in each half of the rope, T ≈ 4,908.8 N.

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2 years ago
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yan [13]

Answer:

g = 1.25m/s²

Explanation:

Given the following data;

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Mathematically, potential energy is given by the formula;

P.E = mgh

Where,

P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

GPE = mgh

Substituting into the equation, we have;

24 = 5*6*g

24 = 30g

g = 30/24

g = 1.25m/s²

Therefore, the acceleration due to gravity on Planet X is 1.25m/s².

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