Physics, 22.06.2019 12:30 clmorcutt420

Apositive charge moves in the direction of an electric field. which of the following statements are true? check all that apply. check all that apply. 1.the potential energy associated with the charge decreases. 2. the electric field does positive work on the charge. 3. the electric field does negative work on the charge. 4. the potential energy associated with the charge increases. 5. the electric field does not do any work on the charge. 6. the amount of work done on the charge cannot be determined without additional information.

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Answer from: Quest

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).

Answer from: Quest

answer: the correct answer id option(d).

explanation:

snell's law refraction:

$\frac{n_2}{n_1}=\frac{sin \theta_1}{sin\theta_2}$

the angle of refraction = $\theta_2=38 ^o$

the angle of incidence = $\theta_1=?$

refractive index of air = $n_1$ = 1

refractive index of glycerol $n_2$ = 1.47

$\frac{1.47}{1}=\frac{sin\theta_1}{sin 38^o}, (sin38^o=0.61)$

$sin\theta_2=0.8967$

$\theta_2=24.2^o, (sin^{-1}0.8967=63.72^o)$

from the given options, the most closest answer to 63.72° is 65° of option (d).

Answer from: Quest

Domain eukarya; kingdom plantae should be the answer.

Answer from: Quest

The figure is missing but we can answer anyway.the correct answer is "releasing the string". in fact, when the person is holding the bow and the arrow, the kinetic energy of the arrow is still zero (because the arrow is not moving). the moment the arrow is released, it acquires a velocity v, so its kinetic energy will become different from zero: $k= \frac{1}{2}mv^2$ where m is the mass of the arrow and v its velocity. all the other options, instead, do not increase the kinetic energy of the system: - pulling farther back on the string : this will just increase the elastic potential energy of the system- holding the arrow higher: this will increase the gravitational potential energy of the system (which is given by u=mgh, where h is the height above the ground)- decreasing the mass of the arrow: this will decrease the gravitational potential energy of the system (which is given by u=mgh, where m is the mass)

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Apositive charge moves in the direction of an electric field. which of the following statements are...

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