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3 years ago

8

Match the given equation with the verbal description of the surface: A. Circular Cylinder B. Plane C. Cone D. Half plane E. Sphe

re F. Elliptic or Circular Paraboloid

1 answer:

sdas [7]3 years ago

8 0

Answer:

A. Circular Cylinder

r=4

$\sqrt{x^{2}+y^{2} }$ =4 independent from z

B. Plane

ρ cos φ ( )=4

spherical coordinates

z=4

C. Cone

φ = π/3

φ spherical coordinates

D. Half Plane

θ = π/3

Cylindrical Coordinates, do not have portion in 3rd quadrant that why its only half a plane.

E. Sphere:

ρ =2cos φ ( )

Spherical Coordinates

ρ =2×z/ρ

$p^{2}$ =2z

$x^{2} +y^{2} +z^{2}$ = $p^{2}$

$x^{2} +y^{2}+z-1^{2}$ =1

F. Elliptic or Circular Parabola

z= $r^{2}$

z= $r^{2}$ = $x^{2} +y^{2}$

z= $y^{2}$

note: pics are attached

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3 years ago

Find the following answer based on the image.

saul85 [17]

<u>We are given:</u>

Mass of Neptune = 1.03 * 10²⁶ kg

Distance from the center of Neptune (r) = 2.27 * 10⁷

now, computing the value of the acceleration due to gravity (g)

<u>Finding g:</u>

We know the formula:

g = G(mass of planet) / (r)²

g = [6.67 * 10⁻¹¹ * 1.03*10²⁶] / (2.27*10⁷) [since G is 6.67*10⁻¹¹]

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which can be rewritten as:

g = (6.87 * 10¹⁵ * 10⁻¹⁴) / 5.15

g = (6.87 * 10¹⁵⁻¹⁴) / 5.15

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3 years ago

In a physics lab, light with a wavelength of 530 nm travels in air from a laser to a photocell in a time of 16.7 ns . When a sla

Harlamova29_29 [7]

Answer:

$\lambda'=78.086\ nm$

Explanation:

Given:

wavelength of light in the air, $\lambda=530\times 10^{-9}\ m$

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time taken to reach the photocell via air and glass slab, $t'=21.3\times 10^{-9}\ s$

thickness of the glass slab, $x=0.87\ m$

<u>Now we have the relation for time:</u>

$\rm time=\frac{distance}{speed}$

hence,

$t=\frac{d}{c}$

c= speed of light in air

$16.7\times 10^{-9}=\frac{d}{3\times 10^8}$

$d=16.7\times 10^{-9}\times 3\times 10^8$

$d=5.01\ m$

For the case when glass slab is inserted between the path of light:

$\frac{(d-x)}{c} +\frac{x}{v} =t'$ (since light travel with the speed c only in the air)

here:

v = speed of light in the glass

$\frac{(5.01-0.87)}{3\times 10^8} +\frac{0.87}{v} =21.3\times 10^{-9}$

$v=4.42\times 10^7\ m.s^{-1}$

Using Snell's law:

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3 years ago

1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp

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Answer with explanation:

We are given that

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Horizontal velocity, $v_x=2 m/s$

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Where $g=9.8 m/s^2$

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$m_1u_1+m_2u_1=-m_1v_1+m_2v_2$

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$v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s$

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Final energy= $\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2$

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Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

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