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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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Answer:

Both technicians A and B

Explanation:

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8 0
3 years ago
A 2000 kg car experiences a braking force of 10000N and skids to a 14 m stop. What was the speed of the car just before the brak
Fudgin [204]

Answer:

V = 11.83 m/s

Explanation:

Given the following data;

Mass = 2000 kg

Force = 10000N

Distance = 14 m

To find the final velocity of the car;

First of all, we would determine the acceleration of the car;

Acceleration = force/mass

Acceleration = 10000/2000

Acceleration = 5 m/s²

Next, we would use the third equation of motion to find the final velocity;

V^{2} = U^{2} + 2aS

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

S represents the displacement measured in meters.

Substituting into the equation, we have;

V² = 0² + 2*5*14

V² = 0 + 140

V = √140

V = 11.83 m/s

5 0
3 years ago
A 120 kg tackler moving at 3.0 m/s meets head-on (and tackles) a 91 kg halfback moving at 7.5 m/s. What will be their mutual vel
Vesna [10]

Explanation:

It is given that,

Mass of the tackler, m₁ = 120 kg

Velocity of tackler, u₁ = 3 m/s

Mass, m₂ = 91 kg

Velocity, u₂ = -7.5 m/s

We need to find the mutual velocity immediately the collision. It is the case of inelastic collision such that,

v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}

v=\dfrac{120\ kg\times 3\ m/s+91\ kg\times (-7.5\ m/s)}{120\ kg+91\ kg}

v = -1.5 m/s

Hence, their mutual velocity after the collision is 1.5 m/s and it is moving in the same direction as the halfback was moving initially. Hence, this is the required solution.

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3 years ago
A student says that a speed of 50 m/s is faster than a speed of 140 km/h because the number is bigger. What would you say to the
ladessa [460]

The units are not consistent - 1 m/s is not the same as 1 km/h.

First thing to do would be to convert from one unit of speed to the other, say km/h to m/s. There are 1000 meters (m) for every kilometer (km) and 3600 seconds (s) for every hour (h), so

1\,\dfrac{\mathrm{km}}{\mathrm h}\cdot\dfrac{10^3\,\mathrm m}{\mathrm{km}}\cdot\dfrac{\mathrm h}{3600\,\mathrm s}\approx0.278\,\dfrac{\mathrm m}{\mathrm s}

So in fact 1 km/h is about 4 times slower than 1 m/s.

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Photographs of many young stars show long jets of material apparently being ejected from their poles.
Rudiy27

Answer:

A) True

Explanation:

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