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liberstina [14]
3 years ago
6

Compare the weight of a mountain climber when she is at the bottom of a mountain with her weight when she is at the top of the m

ountain. In which case is her weight larger? a. Both are the same. b. She weighs four times as much at the top. c. She weighs twice as much at the top. d. She weighs more at the top. e. She weighs more at the bottom.
Physics
1 answer:
kakasveta [241]3 years ago
4 0

Answer:

The correct answer is  a. Both are the same

Explanation:

For this calculation we must use the gravitational attraction equation

    F = G m M / r²

Where M will use the mass of the Earth, m the mass of the girl and r is the distance of the girl to the center of the earth that we consider spherical

To better visualize things, let's repair the equation a little

     F = m (G M / r²)

The amount in parentheses called acceleration of gravity, entered the force called peos

     g = G M / r²

     F = W

    W = m g

When analyzing this equation we see that the variation in the weight of the girl depends on the distance, which is the radius of the earth plus the height where the girl is

    r = Re + h

    Re = 6.37 10⁶ m

    r² = (Re + h)²

    r² = Re² (1 + h / Re)²

Let's replace

    W = m (GM / Re²)   (1+ h / Re)⁻²

    W = m g   (1+ h / Re)⁻²

This is the exact expression for weight change with height, but let's look at its values ​​for some reasonable heights h = 6300 m (very high mountain)

     h / Re = 10 ⁻³

     (1+ h / Re)⁻² = 0.999⁻²

Therefore, the negligible weight reduction, therefore, for practical purposes the weight does not change with the height of the mountain on Earth

The correct answer is a

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When hard stabilization structures such as groins are used to stabilize a shoreline, the change in the longshore current results <u>deposition of sediment. </u>

On the upcurrent side of the barrier, sediment is deposited as the longshore current slows.

What is Hard stabilization?

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4 0
2 years ago
A uranium-238 atom can break up into a thorium-234 atom and a particle called an alpha particle, αα-4. The numbers indicate the
alexdok [17]

Answer: E = 5.80*10^-13 J

Explanation:

Given

We use the law of conservation of momentum to solve this

Momentum before breakup = momentum after breakup

0 = m1v1 + m2v2

0 = 238m * -2.2*10^5 + 4m * v2

0 = -523.6m m/s + 4m * v2

v2 * 4m = 523.6m m/s

v2 = 523.6 m m/s / 4m

v2 = 130.9*10^5 m/s

v2 = 1.31*10^7 m/s

Using this speed in the energy equation, we have

E = 1/2m1v1² + 1/2m2v2²

E = 1/2 * (238 * 1.66*10^-27) * -2.2*10^5² + 1/2 * (4 * 1.66*10^-27) * 1.31*10^7²

E = [1/2 * 3.95*10^-25 * 4.84*10^10] + [1/2 * 6.64*10^-27 * 1.716*10^14]

E = (1/2 * 1.911*10^-14) + (1/2 * 1.139*10^-12)

E = 9.56*10^-15 + 5.7*10^-13

E = 5.80*10^-13 J

3 0
3 years ago
How is the q of an rlc parallel resonant circuit calculated?
notka56 [123]

Answer:

It is calculated by dividing Resistance, R, by Inductive reactance, XL.

Explanation:

Q is called the Q factor of a resonance circuit. In a parallel resonance circuit, it is calculated by finding the ratio of the power stored in the circuit to the power distributed in the circuit. It is a way of measuring the quality of a circuit or how effective the circuit is.

Q factor is the inverse in the resonance series circuit.

Q factor of a resonance parallel circuit,

<h3>Q = R/XL</h3>

R = Resistance

XL = Inductive reactance

3 0
3 years ago
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