Answer:
because he's fast and speedy
Explanation:
Answer:
0.54 A
Explanation:
Parameters given:
Number of turns, N = 15
Area of coil, A = 40 cm² = 0.004 m²
Change in magnetic field, ΔB = 5.1 - 1.5 = 3.6 T
Time interval, Δt = 2 secs
Resistance of the coil, R = 0.2 ohms
To get the magnitude of the current, we have to first find the magnitude of the EMF induced in the coil:
|V| = |(-N * ΔB * A) /Δt)
|V| = | (-15 * 3.6 * 0.004) / 2 |
|V| = 0.108 V
According to Ohm's law:
|V| = |I| * R
|I| = |V| / R
|I| = 0.108 / 0.2
|I| = 0.54 A
The magnitude of the current in the coil of wire is 0.54 A
As ball is projected up in air at an angle of 45 degree without any air resistance
Let the initial speed will be v
now we will have
In x direction
in y direction
now displacement in x direction
displacement in y direction
now from above two equations we have
so above equation is a quadratic equation and hence it will be a parabolic curve
so correct answer will be
<em>C. parabolic curve.</em>
The complete ionization of KBr into its constituents
is:<span>
<span>KBr (s) --->
K+ (aq) + Br- (aq)</span></span>
<span>
During electrolysis, oxidation takes place at the anode electrode. This means
that an ion is stripped off its electron hence becoming more positive:
<span>2 Br- (aq) --->
Br2 (g) + 2e- </span></span>
We can see that Bromine gas Br2 is evolved at the anode.
<span>
<span>Meanwhile at the cathode, the reduction reaction occurs.
Which means that the electron from the anode electrode is used to make an ion
more negative:
<span>2K+ (aq) + 2e- ---> 2K (s) </span></span>
Hence, through reduction, solid potassium is deposited on the
plate.</span>
Half reactions:
<span>Anode: 2 Br- (aq) --->
Br2 (g) + 2e- </span>
<span>Cathode: 2K+ (aq) + 2e-
---> 2K (s) </span>
<span>12-50t=70t, t= 0.1h = 6 minutes.</span>
