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S_A_V [24]
3 years ago
10

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

in meters, and you are standing at a point with coordinates (60, 40, 1966). The positive x-axis points east and the positive y-axis points north. (a) If you walk due south, will you start to ascend or descend? ascend descend Correct: Your answer is correct. At what rate? Correct: Your answer is correct. vertical meters per horizontal meter (b) If you walk northwest, will you start to ascend or descend? ascend descend Correct: Your answer is correct. At what rate? (Round your answer to two decimal places.) Correct: Your answer is correct. vertical meters per horizontal meter (c) In which direction is the slope largest? What is the rate of ascent in that direction? vertical meters per horizontal meter At what angle above the horizontal does the path in that direction begin? (Round your answer to two decimal places.)
Physics
1 answer:
Ierofanga [76]3 years ago
4 0

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

The current location = (60, 40, 1966)  

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

Therefore, walking south increases the elevation, and we ascend

The rate is given by the partial derivative at in the -j direction, which is 0.02

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

Therefore we descend

(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

The slope =(√((-0.4)² + (-0.8)²) = 1

Therefore, the angle is 45° to the horizontal.

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The best way in handling in this situation is that in order for the astronaut to be able to get back to the shuttle is that he or she should take an object from his or her tool belt and to be thrown out away from the shuttle. This will allow her to weight lightly and safely return to the shuttle and would be easier for his or her to do so.
3 0
3 years ago
The inductance in the drawing has a value of L = 9.4 mH. What is the resonant frequency f0 of this circuit?
yuradex [85]

Answer:

The resonant frequency of this circuit is 1190.91 Hz.

Explanation:

Given that,

Inductance, L=9.4\ mH=9.4\times 10^{-3}\ H

Resistance, R = 150 ohms

Capacitance, C=1.9\ \mu F=1.9\times 10^{-6}\ C

At resonance, the capacitive reactance is equal to the inductive reactance such that,

X_C=X_L    

2\pi f_o L=\dfrac{1}{2\pi f_oC}

f is the resonant frequency of this circuit  

f_o=\dfrac{1}{2\pi \sqrt{LC}}

f_o=\dfrac{1}{2\pi \sqrt{9.4\times 10^{-3}\times 1.9\times 10^{-6}}}

f_o=1190.91\ Hz

So, the resonant frequency of this circuit is 1190.91 Hz. Hence, this is the required solution.

4 0
3 years ago
Consider an e. coli to be a cylinder with a diameter of 1 micrometer (um) and with a length of 2 micrometer (um).
julia-pushkina [17]

Answer:

A=6.28\ \mu m^2

Part 1

V=1.57 \ \mu m^3

Part 2

A=6.28\ \mu m^2

Explanation:

Given that

Diameter,d=1 μm

Length ,l=2 μm

As we know that volume of cylinder given as

V=\pi r^2l

V=\pi \times 0.5^2\times 2 \ \mu m^3

V=1.57 \ \mu m^3

Surface area,A

A=π d l

A=\pi \times 1 \times 2\ \mu m^2

A=6.28\ \mu m^2

Part 1

V=1.57 \ \mu m^3

Part 2

A=6.28\ \mu m^2

4 0
3 years ago
An object is thrown vertically upward such that it has a speed of 25 m/s when it reaches two thirds of its maximum height above
AlekseyPX

Answer:

v0 = 25 m/s

vf = 0 m/s

a = -9.80 m/s^2

change in x = 31.89m

but that's only 1/3 of the hight, so i time it by 3 to get 96m

4 0
3 years ago
A cyclist starts from rest and coasts down a 4.0∘ hill. The mass of the cyclist plus bicycle is 85 kg. Ignore air resistance and
Angelina_Jolie [31]

A) change in ht after 180m = 180 * sin(4-deg.) = 12.56m

net work done by gravity on the cyclist = mass * gravity * height diff.

= 85 * 9.8 * 12.56

= 10470J

= 10.5kJ

B) Kinetic energy = 1/2 * mass * vel.^2 = work done by gravity = 10470J

vel.^2 = 10470 * 2 / 85 = 246.4

vel. = 15.7m/s

5 0
3 years ago
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