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STatiana [176]
3 years ago
5

Using the equation for the final velocity in terms of masses and initial velocity of the gliders for a perfectly inelastic colli

sion, work out the final kinetic energy, assuming one of the glider is at rest initially. Compare this to the initial kinetic energy and show that kinetic energy is not conserved in the collision. Do this only with variables; none of your data should be included
Physics
1 answer:
tatuchka [14]3 years ago
4 0

Answer:

Comparison has been made.

Explanation:

Consider perfectly inelastic collision.Two masses of m1 and m2 with initial velocities u1 and u2 before colliding, stick together after collision and move with common speed v. then according to the rule of momentum preservation

m1u1+m2u2=  (m1+m2)v

v= \frac{m1u1+m2u2}{m1+m2}

If u1 and u2 are known the final common velocity can be determined

Initial kinetic energy is

\frac{1}{2}m1u1^{2}+\frac{1}{2}m2u2^{2}

The collion here is perfectly inelastic this means kinectic energy is not conserved. Therefore, initial kinectic energy is not equal to final kinectic.  

\frac{1}{2}m1v^{2}+\frac{1}{2}m2v^{2}

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A flat sheet of paper of area 0.365 m2 is oriented so that the normal to the sheet is at an angle of 60 ∘ to a uniform electric
andriy [413]

Answer:

A.) 3.65 N*m²/C B) No C) 0º D) 90º

Explanation:

A) The electric flux, when the electric field is uniform across a gausssian surface, can be calculated as the dot product of the electric field vector, and the vector representing the area of the surface (normal to the surface and directed outward it by convention), as follows:

Flux = E*A*cos φ

where E = 20 N/C, A = 0.365 m², φ = 60º.

Replacing by the values, we can get the value of the electric flux, as follows:

Flux = 20 N/C* 0.365 m²*0.5 = 3.65 N*m²/C

B) While the area remains constant, and doesn't change orientation, the value of the flux will be the same, regardless the shape of the sheet.

C) When the normal to the sheet and the electric field are parallel each other, the surface will intercept the maximum number of field lines, i.e. the flux will be directly E*A*cos 0º = E*A (maximum value possible).

D) When the electric field is tangent to the surface, this means that no field lines will be intercepted by the sheet, so the flux is zero.

In this case, φ = 90º, cos φ = 0

⇒ E*A*cos 90º = E*A*0 = 0

5 0
3 years ago
Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Give
Keith_Richards [23]

Complete Question

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).

Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Given the satellite specifications listed in the problem introduction, what is the amplitude E0 of the electric field vector of the satellite broadcast as measured at the surface of the earth? Use ϵ0=8.85×10^−12C/(V⋅m) for the permittivity of space and c=3.00×10^8m/s for the speed of light.

Answer:

The electric field vector of the satellite broadcast as measured at the surface of the earth is  E_o = 6.995 *10^{-6} \ V/m

Explanation:

From the question we are told that

     The height of the satellite is  r  = 35000 \ km  =  3.5*10^{7} \ m

      The power output of the satellite is P  = 1 \ KW  =  1000 \ W

       

Generally the intensity of the electromagnetic radiation of the satellite at the surface of the earth is  mathematically represented as  

     I  =  \frac{P}{4 \pi r^2}

substituting values

      I  =  \frac{1000}{4 * 3.142 (3.5*10^{7})^2}

      I  = 6.495*10^{-14} \  W/m^2

This intensity of the electromagnetic radiation of the satellite at the surface of the earth can also be   mathematically represented as  

          I  =  c * \epsilon_o * E_o^2

Where E_o is the amplitude of the electric field vector of the satellite broadcast so

         E_o =  \sqrt{\frac{2 * I}{c * \epsilon _o} }

substituting values

          E_o =  \sqrt{\frac{2 * 6.495 *10^{-14}}{3.0 *10^{8} * 8.85*10^{-12}} }

           E_o = 6.995 *10^{-6} \ V/m

 

   

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Answer:

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