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docker41 [41]
2 years ago
6

Hydrostatic equilibrium in the Sun means that Choose one: A. energy produced in the core per unit time equals energy emitted at

the surface per unit time. B. pressure balances the weight of overlying layers. C. the Sun absorbs and emits equal amounts of energy. D. the Sun does not change over time.
Physics
1 answer:
kati45 [8]2 years ago
8 0

Answer: Hydrostatic equilibrium in the Sun means that pressure balances the weight of overlying layers.

Explanation: To find the answer, we have to know more about the Hydrostatic equilibrium in sun.

<h3>What is Hydrostatic equilibrium in sun?</h3>
  • Hydrostatic equilibrium in sun means that, the sun is neither expanding not contracting.
  • We can say that the forces that control the sun are precisely balanced.
  • The structure of the sun is in such a way that it can balance the pressure outward and the force of gravity inward.
  • This balance is known as hydrostatic equilibrium.

Thus, we can conclude that, the Hydrostatic equilibrium in the Sun means that pressure balances the weight of overlying layers.

Learn more about the hydrostatic equilibrium in sun here:

brainly.com/question/28044769

#SPJ4

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IgorC [24]

Answer: The acceleration of the object is 0.67m/s^2 west.

Explanation: Here we are given the initial velocity and final velocity as well as the time taken. Acceleration is the change in velocity per unit time, thus the equation becomes.

a=dv/t

a=vf-vi/t

a=-2.1-4.7/3.9

a= 0.66m/s^2 west

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A material through which electrons can move easily is a what
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Electrical conductors

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One of the most important properties of materials in many applications is strength. Two of the qualitative measures of the stren
katrin2010 [14]

To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.

In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.

By definition we know that the tensile strength is defined as

\sigma = \frac{F}{A}

Where,

\sigma =Tensile strength

F = Tensile Force

A = Cross-sectional Area

In the other hand we have that the shear strength is defined as

\sigma_y = \frac{F_y}{A}

where,

\sigma_y =Shear strength

F_y = Shear Force

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PART A) Replacing with our values in the equation of tensile strenght, then

311*10^6 = \frac{F}{(15*10^{-6})(30*10^{-2})}

Resolving for F,

F= 1399.5N

PART B) We need here to apply the shear strength equation, then

\sigma_y = \frac{F_y}{A}

210*10^6 = \frac{F_y}{15*10^{-6}30*10^{-2}}

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6 0
3 years ago
A World War II bomber flies horizontally over level terrain, with a speed of 287 m/s relative to the ground and at an altitude o
Scorpion4ik [409]

Answer: 7.38 km

Explanation: The attachment shows the illustration diagram for the question.

The range of the bomb's motion as obtained from the equations of motion,

H = u(y) t + 0.5g(t^2)

U(y) = initial vertical component of velocity = 0 m/s

That means t = √(2H/g)

The horizontal distance covered, R,

R = u(x) t = u(x) √(2H/g)

Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2

R = 287 √(2×3240/9.8) = 7380 m = 7.38 km

6 0
3 years ago
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