Complete question is;
Selena drew a diagram to show how current moves in a loop of wire that is placed between two magnets. At top left a piece of magnet labeled N and at top right a piece labeled S. Between these a square coil of wire with green arrows at its ends away from the magnets pointing away from the magnets. Red arrows point from N to S. Which best explains how Selena can correct her error?
A) She can change the magnets so like poles are facing each other.
B) She can change the arrows so they show current traveling in opposite directions on the sides of the loop.
C) She can change the arrows so they show the magnetic field reversing direction between the magnets.
D) She can change the magnetic poles so they are both on one side of the loop.
Answer:
B) She can change the arrows so they show current traveling in opposite directions on the sides of the loop.
Explanation:
Since the direction of the arrows point north and south whereas the magnets are east and west, it means the arrows are not pointing in the correct direction. To correct it she will have to make sure the arrows are pointing in opposite directions of the magnet via the sides of the loop.
Answer:
The smaller, inner planets include Mercury, Venus, Earth, and Mars. The inner planets are rocky and have diameters of less than 13,000 kilometers. The outer planets include Jupiter, Saturn, Uranus, and Neptune. The outer planets are called gas giants and have a diameter of greater than 48,000 kilometers.
Explanation:
Its probably a graduated cylinder of something in that nature
Answer:
a) When its length is 23 cm, the elastic potential energy of the spring is
0.18 J
b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
Explanation:
Hi there!
a) The elastic potential energy (EPE) is calculated using the following equation:
EPE = 1/2 · k · x²
Where:
k = spring constant.
x = stretched lenght.
Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).
First, let´s convert the spring constant units into N/m:
4 N/cm · 100 cm/m = 400 N/m
EPE = 1/2 · 400 N/m · (0.03 m)²
EPE = 0.18 J
When its length is 23 cm, the elastic potential energy of the spring is 0.18 J
b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:
EPE = 1/2 · 400 N/m · (0.06 m)²
EPE = 0.72 J
When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
