Physics, 26.10.2021 17:40, jasoncarter

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Physics, 21.06.2019 20:20, jacesmokesloud7254

Copper has free electrons per cubic meter. a 71.0-cm length of 12-gauge copper wire that is 2.05 mm in diameter carries 4.85 a of current. (a) how much time does it take for an electron to travel the length of the wire? (b) repeat part (a) for 6-gauge copper wire (diameter 4.12 mm) of the same length that carries the same current. (c) generally speaking, how does changing the diameter of a wire that carries a given amount of current affect the drift velocity of the electrons in the wire?

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Physics, 22.06.2019 01:10, twhalon72

The x-coordinate of a particle in curvilinear motion is given by x = 3.1t3 - 4.9t where x is in feet and t is in seconds. the y-component of acceleration in feet per second squared is given by ay = 2.3t. if the particle has y-components y = 0 and vy = 5.0 ft/sec when t = 0, find the magnitudes of the velocity v and acceleration a when t = 1.8 sec. sketch the path for the first 1.8 seconds of motion, and show the velocity and acceleration vectors for t = 1.8 sec. answers: v = ft/sec a = ft/sec2

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Physics, 22.06.2019 12:10, hilljade45

Two forces produce equal torques on a door about the door hinge. the first force is applied at the midpoint of the door; the second force is applied at the doorknob. both forces are applied perpendicular to the door. which force has a greater magnitude?

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Physics, 22.06.2019 23:00, shelbybibb99

Acommon technique in analysis of scientific data is normalization. the purpose of normalizing data is to eliminate irrelevant constants that can obscure the salient features of the data. the goal of this experiment is to test the hypothesis that the flux of light decreases as the square of the distance from the source. in this case, the absolute value of the voltage measured by the photometer is irrelevant; only the relative value conveys useful information. suppose that in part 2.2.2 of the experiment, students obtain a signal value of 162 mv at a distance of 4 cm and a value of 86 mv at a distance of 5.7 cm. normalize the students' data to the value obtained at 4 cm. (divide the signal value by 162.) then calculate the theoretically expected (normalized) value at 5.7 cm.

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