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3 years ago

As an object rolls downhill, some of the energy is

2 answers:

7 0

When the object is at the top of the hill it has the most potential energy. If it is sitting still, it has no kinetic energy. As the object begins to roll down the hill, it loses potential energy, but gains kinetic energy. The potential energy of the position of the object at the top of the hill is getting converted into kinetic energy. Hope this helped. :)

mr_godi [17]3 years ago

6 0

Well this is kind of a trick question. When the object is still before it rolls downhill it is potential. But as the object is rolling down the hill, it is kinetic energy. So, I believe your answer would be kinetic since it doesn't specify everything that's happening.

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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

Katyanochek1 [597]

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;

diameter of each ring, d = 10 cm = 0.1 m

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;

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

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

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3 years ago

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Dahasolnce [82]

Answer:

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2 years ago

Read 2 more answers

JOSEPH JOGS FROM END A TO OTHER END B OF A straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back

zzz [600]

(a) The average speed from A to B would be 1.76 metre per second and the average velocity from A to B would also be 1.76 metre per second

<span>(b) The average speed from A to C would be 1.73 metre per second and the average velocity from A to C would be 0.87 metre per second</span>

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3 years ago

A swimmer, capable of swimming at a speed of 1.0 m/s in still water (i.e., the swimmer can swim with a speed of 1.0 m/s relative

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If the current takes him downstream we must find the resultant vector of the velocities: $V res= \sqrt{1^{2}+0.91^{2} } = \sqrt{1.8281}= 1.3520747$ Then if the river is 3000 m-wide the swimmer will have to pass:
1.3520747 · 300 = 4056.14 m t = 4056.14 m : 1 m/s
a ) It takes 4056.15 seconds ( 1 hour 7 minutes and 36 seconds ) to cross the river.
b ) 0.91 · 3000 = 2730 m
He will be 2730 m downstream.

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3 years ago

How far will a man travel in 15 min driving his car down the highway at an average speed of 24 m/?s

emmasim [6.3K]

Answer:

21.6 km

Explanation:

15×60×24 = 21600 m

$\frac{21600}{1000}$ = 21.6 km

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3 years ago