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Kamila [148]
3 years ago
7

A student wants to investigate the motion of a ball by conducting two different experiments, as shown in Figure 1 and Figure 2 a

bove. In Experiment 1, the student releases a ball from rest and uses a slow-motion camera to film the ball as it falls to the ground. Using video analysis, the student is able to plot the ball’s horizontal position x and vertical position y as a function of time t. In Experiment 2, the student horizontally rolls the same ball off a table, and uses video analysis to plot the ball’s horizontal position x and vertical position y as a function of time t starting from the instant the ball leaves the table. The graphs from each experiment are shown above along with each graphs best-fit curve line. In Experiment 1, what is the speed of the ball the instant it makes contact with the ground?
Physics
1 answer:
Lelu [443]3 years ago
5 0
It equals to G



^the answer
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Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field
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Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength  at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

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distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength  at the midpoint between the two rings is given as;

E_{mid} = E_{right} +E_{left}\\\\E_{right}  = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt}  = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }  - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0

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