Answer:
25.08m/s
Explanation:
mgh1 + 0.5mv1² = mgh2 + 0.5mv2²
h1 = 0m
v1 = u
h2 = 5m
v2 = 23m/s
putting the values into the formula above;
m(10)(0) + 0.5m(u²) = m(10)(5) + 0.5m(23²)
0 + 0.5mu² = 50m + 264.5m
0.5mu² = 314.5m
dividing through by m
0.5u² = 314.5
u² = 629
u = <u>2</u><u>5</u><u>.</u><u>0</u><u>8</u><u>m</u><u>/</u><u>s</u>
<u>Theref</u><u>ore</u><u>,</u><u> </u><u>the</u><u> </u><u>init</u><u>ial</u><u> </u><u>speed</u><u> </u><u>"</u><u>u</u><u>"</u><u> </u><u>=</u><u> </u><u>2</u><u>5</u><u>m</u><u>/</u><u>s</u>
KE = 1/ 2 * 1252 * 144
as KE = 1/2 * m * v ^2
= 90144 J
Answer:
The net torque is zero
Explanation:
Let's assume that the dipole is compose of two equal but oposite charges e, and it cam be represented by a rod with one end having a charge e and the other end with a charge of -e. Notice that the dipole is parallel to the electric field thus the force felt by both of the charges will be parallel to the electric field. This means that there will be no components of the forces that are perpendicular to the rod which is a requirement for it to have a torque.
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<span><span>The
best and most correct answer among the choices provided by the question is </span>B.-2.71 V.</span>
Mg2+(aq) + 2e- -> Mg(s) E=-2.37 V
Cu2+(aq) + 2e- -> Cu(s) E =+ 0.34 V
Since Cu is acting as the anode, the equation needs to be
reversed.
Cu(s) -> Cu2+(aq) + 2e- E =- 0.34 V
Ecell= -2.37 V+ (- 0.34 V) = -2.71 V
<span><span>
</span><span>Hope my answer would be a great help for you. </span> </span>
<span> </span>
Resistance per 1000 feet for gauge 14 wire is given as
R = 2.525 ohm
now if wire is of length 50 feet only then the resistance is given as


now if 11 A current flows through the wire then the voltage drop is given by ohm's law



so most appropriate answer in given options is
A. 1.8 Volts