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Ugo [173]
3 years ago
15

An 85-kg hiker climbs to the summit of Mount Mitchell in western North Carolina. During one 2.0-h period, the climber's vertical

elevation increases 540 m. Determine the change in gravitational potential energy of the climber-Earth system.
Physics
1 answer:
Bas_tet [7]3 years ago
5 0

Answer:

The change in gravitational potential energy of the climber-Earth system,\Delta E=4.49\times 10^5\ J

Explanation:

Given that,

Mass of the hiker, m = 85 kg

Time, t = 2 h

Vertical elevation of the climber, h = 540 m

We need to find the change in gravitational potential energy of the climber-Earth system. We know that due to change in position of an object, gravitational potential energy occurs. It is given by :

\Delta E=mgh\\\\\Delta E=85\times 9.8\times 540\\\\\Delta E=4.49\times 10^5\ J

So, the change in gravitational potential energy of the climber-Earth system is 4.49\times 10^5\ J. Hence, this is the required solution.  

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MGS and later orbiters found spectral evidence for what minerals on the Martian surface? Check all that apply.
slamgirl [31]

Mars Global Surveyors (MGS) and later orbiters found the following  minerals on the  Martian surface;

  • Phyllosilicates
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The Mars Global Surveyors (MGS) and later orbiters suggest that the Martian crust contains a higher percentage of volatile elements such as Sulphur and chlorine than the Earth's crust does.

These scientists also conclude that the most abundant chemical elements  in the Martian crust are those found in Igneous rock.

These elements include the following;

  • Silicon,
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They also, suggest that hydrogen is found in ice (water) while carbon is found in carbon dioxide and carbonates.

From the given options the minerals found in Martian surface include;

  • Phyllosilicates ------ these are sheet of silicate minerals
  • Carbonate
  • Sulfates
  • iron oxide

Learn more here: brainly.com/question/20470323

6 0
2 years ago
the carnot cycle attempts to model the most efficient possible process by avoiding what? adiabatic processes isothermal processe
kobusy [5.1K]

The carnot cycle attempts to model the most efficient possible process by avoiding irreversible processes.

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Essentially, this means that any reversible cycle can be performed in reverse and that the amount of work or heat exchanged along the forward and backward pathways is the same.

It goes without saying that such reversible processes are not possible because they would take an unlimited amount of time. Therefore, the Carnot Engine is described as an idealized heat engine that uses the Carnot Cycle, a reversible cycle.

Learn more about carnot cycle here;

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5 0
7 months ago
During a hurricane in 2008, the Westin Hotel in downtownNew Orleans suffered damage. Suppose a piece of glass dropped near the t
tangare [24]

Answer:

  • <u>77.8 m/s, downward</u>

Explanation:

For uniform acceleration motion, the average speed is equal to half the soum of the initial velocity, Vi, and the final velocity, Vf

  • Average speed = (Vf + Vi)/2

Also, by definition, the average speed is the distance divided by the time:

  • Average speed = distance / time

Then:

  • (Vf + Vi)/2 = 300m/6.62s

Other kinematic equation for uniform acceleration is:

  • Vf = Vi + a×t

Since the window is falling and the air resistance is ignored, a = g (gravitational acceleration ≈ 9.8m/s²)

Replacing the known values we can set a system of two equations:

From (Vf + Vi)/2 = 300m/6.62s

(Vf + Vi) = 2 × 300m/6.62s

  • Vf + Vi = 90.634      equation 1

From Vf = Vi + a×t

Vf - Vi = 9.8 (6.62)

  • Vf - Vi = 64.876     equation 2

Adding the two equations:

  • 2Vf = 155.510

  • Vf = 77.8 m/s downward (velocities must be reported with their directions)
8 0
2 years ago
Sound produced when sections of a string vibrate separately are called
sergiy2304 [10]
Lungs vacoules on if those 2
5 0
2 years ago
If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g
Bas_tet [7]

Answer:

  W = 1,307 10⁶ J

Explanation:

Work is the product of force by distance, in this case it is the force of gravitational attraction between the moon (M) and the capsule (m₁)

              F = G m₁ M / r²

              W = ∫ F. dr

              W = G m₁ M ∫ dr / r²

we integrate

             W = G m₁ M (-1 / r)

                 

We evaluate between the limits, lower r = R_ Moon and r = ∞

           W = -G m₁ M (1 /∞ - 1 / R_moon)

            W = G m1 M / r_moon

Body weight is

             W = mg

             m = W / g

The mass is constant, so we can find it with the initial data

For the capsule

            m = 1000/32 = 165 / g_moon

            g_moom = 165 32/1000

            .g_moon = 5.28 ft / s²

I think it is easier to follow the exercise in SI system  

           W_capsule = 1000 pound (1 kg / 2.20 pounds)

           W_capsule = 454 N

           W = m_capsule g

           m_capsule = W / g

           m = 454 /9.8

           m_capsule = 46,327 kg

Let's calculate

          W = 6.67 10⁻¹¹ 46,327   7.36 10²² / 1.74 10⁶

          W = 1,307 10⁶ J

7 0
3 years ago
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