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alukav5142 [94]
3 years ago
14

A 66.5-kg hiker starts at an elevation of 1270 m and climbs to the top of a peak 2660 m high.

Physics
1 answer:
anyanavicka [17]3 years ago
8 0

Answer

given,

mass of the hiker = 66.5 kg

elevation of the hiker,h₁ = 1270 m

elevation of top peak,h₂ = 2660 m

a) Change in potential energy

    Δ PE = m g h₂ - m g h₁

    Δ PE = m g (h₂ - h₁)

    Δ PE = 66.5 x 9.8 x  (2660 - 1270)

    Δ PE = 905863 J

b) Minimum work require by the hiker will be equal to Δ PE = 905863 J

c) yes, it can be more than this, if friction is present on the surface.

   

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Answer:

The strong attraction of each shared electron to both nuclei stabilizes the system, and the potential energy decreases as the bond distance decreases. If the atoms continue to approach each other, the positive charges in the two nuclei begin to repel each other, and the potential energy increases.

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3 years ago
A 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a very light rope that goes over an ideal pulley. If the
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Answer:

acceleration = 2.4525‬ m/s²

Explanation:

Data: Let m1 = 3.0 Kg, m2 = 5.0 Kg, g = 9.81 m/s²

Tension in the rope = T

Sol: m2 > m1

i) for downward motion of m2:

m2 a = m2 g - T

5 a = 5 × 9.81 m/s² - T  

⇒ T = 49.05‬ m/s² - 5 a     Eqn (a)‬

ii) for upward motion of m1

m a = T - m1 g

3 a = T - 3 × 9.8 m/s²

⇒ T =  3 a + 29.43‬ m/s²   Eqn (b)

Equating Eqn (a) and(b)

49.05‬ m/s² - 5 a = T =  3 a + 29.43‬ m/s²

49.05‬ m/s² - 29.43‬ m/s² = 3 a + 5 a

19.62 m/s² = 8 a

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. A car starts to move from rest. If its velocity becomes 90km/hr after 3s, calculate its acceleration​
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32.5/s squared

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3 years ago
If Vx = 7.00 units and Vy = -7.60 units, determine the magnitude of V⃗ .
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|V| = 10.33 units and the direction θ = -47.35° or 312.65°.

Given the x and y components of a vector, we can calculate the magnitude and direction from these components.

Applying the Pythagorean theorem we have that the magnitude of the vector is:

|V| = \sqrt{Vx^{2}+Vy^{2}  }

|V| = \sqrt{(7.00units)^{2}+(-7.60units)^{2}} = \sqrt{106units^{2}} = 10.33units

The expression for the direction of a vector comes from the definition of the tangent of an angle:

tan θ = \frac{Vy}{Vx} ------>  θ = arc tan \frac{Vy}{Vx}

θ = arc tan \frac{-7.60units}{7.00units}

θ = -47.35° or 312.65°

6 0
3 years ago
A fire truck drives 3 miles to the east then 1.2 miles to west what is the displacement of the truck
Rudiy27

The answer to the question you have asked is 1.8 miles east.

8 0
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