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myrzilka [38]
3 years ago
9

At 45° latitude, the gravitational acceleration as a function of elevation z above sea level is given by g = a − bz , where a =

9.807 m/s2 and b = 3.32 × 10−6 s−2 . Determine the height above sea level where the weight of an object will decrease by 0.3 percent. Answer: 8862 m
Engineering
1 answer:
Ahat [919]3 years ago
7 0

Answer:

8861.75 m approximately 8862 m

Explanation:

We need to remember Newton's 2nd Law which says that the force experienced by an object is proportional to his acceleration and that the constant of proportionality between those two vectors correspond to the mass of the object.

F=ma for the weight of an object (which is a force) we have that the acceleration experienced by that object is equal to the gravitational acceleration, obtaining that  W = mg

For simplicity we work with g =9.807 \frac{m}{s^{2}} despiting the effect of the height above sea level. In this problem, we've been asked by the height above sea level that makes the weight of an object 0.30% more lighter.

In accord with the formula g = a-bz the "normal" or "standard" weight of an object is given by W = mg = ma when z = 0, so we need to find the value of z that makes W = m(a-bz) = 0.997ma meaning that the original weight decrease by a 0.30%, so now we operate...

m(a-bz) = 0.997ma now we group like terms on the same sides ma(1-0.997) = mbz we cancel equal tems on both sides and obtain that z = \frac{a}{b} (0.003) = \frac{9.807 \frac{m}{s^{2} } }{3.32*10^{-6} s^{-2} } (0.003) = 8861.75 m

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Answer:

In the center and directed away from it.

Explanation:

The direction along the radius and directed away from the center is known as radial direction.

The velocity is highest in the radial direction pointing away from the center, this is because of the reason that  when the particle executes its motion in the direction that is radial, then it is not acted upon by any force that opposes the motion of the particle and thus there is no obstruction to the velocity of the particle and it is therefore, the highest in the radial direction.

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3 years ago
A man weighs 145 lb on earth.Part ASpecify his mass in slugs.Express your answer to three significant figures and include the ap
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Answer:

<em>a) 4.51 lbf-s^2/ft</em>

<em>b) 65.8 kg</em>

<em>c) 645 N</em>

<em>d) 23.8 lb</em>

<em>e) 65.8 kg</em>

<em></em>

Explanation:

Weight of the man on Earth = 145 lb

a) Mass in slug is...

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b) Mass in kg is...

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x = 145/2.205 = <em>65.8 kg</em>

c) Weight in Newton = mg

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m is mass in kg

g is acceleration due to gravity on Earth = 9.81 m/s^2

Weight in Newton = 65.8 x 9.81 = <em>645 N</em>

d) If on the moon with acceleration due to gravity of 5.30 ft/s^2,

1 m/s^2 = 3.2808 ft/s^2

x m/s^2 = 5.30 ft/s^2

x = 5.30/3.2808 = 1.6155 m/s^2

weight in Newton = mg = 65.8 x 1.6155 = 106

weight in pounds = 106/4.448 = <em>23.8 lb</em>

e) The mass of the man does not change on the moon. It will therefore have the same value as his mass here on Earth

mass on the moon = <em>65.8 kg</em>

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Water at a pressure of 3 bars enters a short horizontal convergent channel at 3.5 m/s. The upstream and downstream diameters of
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Answer:

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Explanation:

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\frac{P_{u}}{\gamma _{w}}+\frac{V-{u}^{2}}{2g}+z_{u}=\frac{P_{L}}{\gamma _{w}}+\frac{V{L}^{2}}{2g}+z_{L}

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Applying the values in the Bernoulli's equation we get

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