Answer:
U = 102.8 J (100 J to two significant digits)
Explanation:
potential energy converted = 20(9.8)(1.8) = 352.8 J
kinetic energy at base of track = ½(20)5.0² = 250 J
energy (work) of friction 352.8 - 250 = 102.8 J
<span>You are given a QL = -26 μC charge that is placed on the x-axis at x = - 0.2 m and a QR = 26 μC charge that is placed at x = +0.2 m. The answers are:
The x-component of the electric field at x = 0 m and y = 0.2 m is 3.
The y-component of the electric field at x = 0 m and y = 0.2 m is 2.
</span>
Answer:
98 joules
Explanation:
1/2 of 28 = 14
14 multiply by 7= 98 joules
Answer:
the ionosphere
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Answer:B. Increased the amount of charge.
Explanation:
