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Dovator [93]
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

Physics
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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The zero net electric field point is at a point that is 0.98 m away from 4.7C charge.If a 14C charge is placed at this point then, force acted on the charge placed at this point is equal to zero.

Explanation:

Let at A both net electric field is zero then

At A ,E1=E2

E1=k*Iq1I / (d+x)^2

E2=k*Iq2I /x^2

Equating both

6 0
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]

g = (6.87 * 10¹⁵) / (5.15 * 10¹⁴)

which can be rewritten as:

g = (6.87 * 10¹⁵ * 10⁻¹⁴) / 5.15

g = (6.87 * 10¹⁵⁻¹⁴) / 5.15

g = (6.87/5.15) * 10

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5 0
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
  • time taken to travel from the source to the photocell via air, t=16.7\ s
  • 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:

\frac{\lambda}{\lambda'} =\frac{c}{v}

\frac{530}{\lambda'} =\frac{3\times 10^8}{4.42\times 10^7}

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5 0
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
Tanzania [10]

Answer with explanation:

We are given that  

Mass of ball,m_1=75 g=\frac{75}{1000}=0075kg

1 kg=1000 g

Height,h_1=1.6 m

h_2=0.6 m

Horizontal velocity,v_x=2 m/s

Mass of platem_2=400 g=\frac{400}{1000}=0.4 kg

a.Initial velocity of plate,u_2=0

Velocity before impact=u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s

Where g=9.8 m/s^2

Velocity after impact,v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s

According to law of conservation of momentum  

m_1u_1+m_2u_1=-m_1v_1+m_2v_2

Substitute the values  

0.075\times 5.6+0=-0.075\times 3.4+0.4v_2

0.4v_2=0.075\times 5.6+0.075\times 3.4

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

Velocity of plate=1.69 m/s

b.Initial energy=\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J

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

Final energy=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J

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

6 0
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