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Anna71 [15]
2 years ago
6

Which of the following are true statements about the electric field? a. A moving charge produces a magnet anomaly. b. The electr

ic field points towards a negative charge and away from a positive charge. c. The direction of the electric field is the direction of acceleration of a positive test charge. d. The electric field is a charge-specific force field independent of any test charge. e. Equipotential lines imply magnitude and direction of the electric field. f. The electric field is a vector field that is found by taking the gradient of the negative potential. g. The electric field inside a spherical conducting shell in 3D is zero. h. If the electric field inside a conductor is nonzero, charges must not be moving; that is, there is a zero current. i. No electric field exists between the plates of a charged capacitor. j. The electric lines of force indicate the vector force experienced by a positive test charge released at a particular point.
Physics
1 answer:
mars1129 [50]2 years ago
4 0

Answer:

a,b,c,d,,f, g, j

Explanation:

e) equipotential lines are lines connecting points of equal potential

h) electric field inside the conductor is non-zero even when there is net movement of charge or non-zero current.

i) capacitors' plates are charged and an electric field exists between the plates.

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If part of an electric circuit dissipates energy at 6 W when it draws a current of 3 A, what voltage is impressed across it?
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To understand how to find the velocities of objects after a collision.
trasher [3.6K]

There are some information missing on Part D: Let the mass of object 1 be m and the mass of object 2 be 3m. If the collision is perfectly inelastic, what are the velocities of the two objects after the collision? Give the velocity v_1 of object one, followed by object v_2 of object two, separated by a comma. Express each velocity in terms of v.

Answer: Part A: v_1 = 0; v_2 = v

Part B: v_1 = v_2 = \frac{v}{2}

Part C: v_1 = \frac{v}{3}; v_2 = \frac{4v}{3}

Part D: v_1 = v_2 = \frac{v}{4}

Explanation: In elastic collisions, there no loss of kinetic energy and momentum is conserved. Momentum is determined as p = m.v and kinetic energy as K = \frac{1}{2}m.v^{2}

Conserved means that the amount of initial momentum is equal to the amount of final momentum:

m_{1}.v_{1i} + m_{2}.v_{2i} = m_{1}.v_{1f} + m_{2}.v_{2f}

No loss of energy means that initial kinietc energy is the same as the final kinetic energy:

\frac{1}{2}(m_{1}.v_{1i} + m_{2}.v_{2i}) = \frac{1}{2} (m_{1}.v_{1f} + m_{2}.v_{2f}  )

To determine the final velocities of each object, there are 2 variables and two equations, so working those equations, the result is:

v_{2f} = \frac{2.m_{1} } {m_{1} + m_{2} }.v_{1i}  + \frac{(m_{2} - m_{1})}{m_{1} + m_{2} } . v_{2i}

v_{1f} = \frac{m_{2} - m_{1} }{m_{1} + m_{2} } . v_{1i} + \frac{2.m_{2} }{m_{1} + m_{2} } .v_{2i}

For all the collisions, object 2 is static, i.e. v_{2i} = 0

<u>Part A</u>: Both objects have the same mass (m), v_{1i} = v and collision is elastic:

v_1 = \frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}

v_1 = 0

v_2 = \frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}

v_2 = \frac{2.m}{m+m}.v

v_2 = v

When the masses are the same and there is an object at rest, the object in movement stops and the object at rest has the same same velocity as the object who hit it.

<u>Part B</u>: Same mass but collision is inelastic: An inelastic collision means that after it happens, the two objects has the same final velocity, then:

m_{1}.v_{1i} + m_{2}.v_{2i} = m_{1}.v_{1f} + m_{2}.v_{2f}

m_{1}.v_{1i} = (m_{1}+m_{2}).v_{f}

v_{f} =  \frac{m_{1}.v_{1i}}{m_{1} + m_{2} }

v_1 = v_2 = \frac{m.v}{m+m}

v_1 = v_2 = \frac{v}{2}

<u>Part C:</u> Object 1 is 2m, object 2 is m and elastic collision:

v_1 = \frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}

v_1 = \frac{2m - m}{2m + m } . v

v_1 = \frac{v}{3}

v_2 = \frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}

v_2 = \frac{2.2m}{2m+m}.v

v_2 = \frac{4v}{3}

<u>Part D</u>: Object 1 is m, object is 3m and collision is inelastic:

v_1 = v_2 = v_{f} =  \frac{m_{1}.v_{1i}}{m_{1} + m_{2} }

v_1 = v_2 = \frac{m}{m+3m}.v

v_1 = v_2 = \frac{v}{4}

5 0
3 years ago
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