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shutvik [7]
3 years ago
7

10. Suppose the tern travels 1.70*10^ km south, only to encounter bad weatherInstead of trying to fly around the stormthe tem tu

ms around and travels 6.00 * 10 ^ 2 10'north wait out the stormIt then tums around again immediately and flies 144 to Antarctica What are the average speed and velocity if it makes this in 122 days ?

Physics
1 answer:
Ghella [55]3 years ago
3 0

We can define average speed as the quotient between the whole distance traveled and the time it takes to travel that distance, and the velocity as the quotient between the displacement and the time it takes to displace that quantity.

We will find that:

Average speed =  2.6×10^2 km/day

Velocity = 2.52×10^2 km/day

So we know that:

The tent travels:

1.70×10^4 km to the south

6.00×10^2 km to the north

1.44×10^4 km to the south.

The total distance traveled is just the sum of the 3 above distances:

total distance = 1.70×10^4 km +  6.00×10^2 km + 1.44×10^4 km

Notice that the second term has a different exponent than the others, so to have the same exponent we can write:

6.00×10^2 km = 0.06×10^4 km

Now we get:

total distance = 1.70×10^4 km + 0.06×10^4 km + 1.44×10^4 km

total distance = (1.70 + 0.06 + 1.44)×10^4 km

<u>total distance = 3.2×10^4 km</u>

And it travels that distance in 122 days, then the average speed is:

AS = (3.2×10^4 km)/(122 days) = 2.6×10^2 km/day

Now we need to find the displacement, which is the difference between the final position and the initial position, the displacement will just be:

d =  1.70×10^4 km - 0.06×10^4 km + 1.44×10^4 km

(We add the distance traveled to south and subtract the distance that it travels due north)

d = (1.70 - 0.06 + 1.44)×10^4 km = 3.08×10^4 km

Then the velocity will be:

v = (3.08×10^4 km)/(122 days) = 2.52×10^2 km/day

Because we defined south as positive and north as negative, having a positive velocity means that the direction of the <u>velocity is due south.</u>

If you want to learn more, you can read:

brainly.com/question/12322912

 

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Suppose you are climbing a hill whose shape is given by the equation z = 2000 − 0.005x2 − 0.01y2, where x, y, and z are measured
Ierofanga [76]

Answer:

(a) Ascend at 0.8 vertical meter/meter

(b) Descend at -0.2·√2 vertical meter/meter

(c) In the (-0.6, -0.8) direction. The path begins at 45° to the horizontal

Explanation:

The given equation of the shape of the hill is z = 2000 - 0.005·x² - 0.01·y²

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The direction of the positive x-axis = east

The direction of the positive y-axis = north

(a) Walking due south = Reducing the y-value 40

From the equation, the elevation varies inversely with the motion towards the north

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The rate is therefore 40 × 0.02 = 0.8

(b)The unit vector in the northwest direction u = 1/√2·(-1, 1)

∴ The rate = (-0.01(60), -0.02(40))·u = (-0.6, -0.8)·1/√2·(-1, 1) = -0.2·√2

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(c) The slope is largest in the grad of the function at the point (60, 40) which is given as follows;

d(2000 - 0.005·x² - 0.01·y² )/dx, d(2000 - 0.005·x² - 0.01·y² )/dy = (-0.6, -0.8)

Therefore, the direction is tan⁻¹(-0.8/-0.6) ≈ S 36.87° W

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Alecsey [184]
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On the surface of the Earth,

                   Weight = (60 kg) x (9.8 m/s²)

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Now, the force of gravity varies as the inverse of the square of the distance from the center of the Earth.
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Again, the object's mass is still 60 kg out there.
___________________________________________

If you have a textbook, or handout material, or a lesson DVD,
or a teacher, or an on-line unit, that says the object "weighs"
60 kilograms, then you should be raising a holy stink. 
You are being planted with sloppy, inaccurate, misleading
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They owe you better material.
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