Answer:
Explanation:
We shall apply conservation of mechanical energy
kinetic energy of alpha particle is converted into electric potential energy.
1/2 mv² = k q₁q₂/d , d is closest distance
d = 2kq₁q₂ / mv²
= 2 x 9 x 10⁹ x 79e x 2e / 4mv²
= 1422 x2x (1.6 x 10⁻¹⁹)² x 10⁹ /4x 1.67 x 10⁻²⁷ x (1.5 x 10⁷)²
= 3640.32 x 10⁻²⁹ /2x 3.7575 x 10⁻¹³
= 484.4 x 10⁻¹⁶
=48.4 x 10⁻¹⁵ m
The answer is D. Electric resistance increases with an increase in the length of a wire and as a result current flow decreases. There is a direct relationship between the length of the wire and the resistance. The longer the wire, the more resistance there will be. Additionally, from Ohm's Law, current is inversely proportional to resistance. This means as the current increases, resistance decreases or vice versa.
... find length
(way 1) determine acceleration using force
only force act on skier is mg vertically. spilt vector we get force along the incline = mgsin(10) and f= ma so
ma = mgsin(10) or a = gsin(10)
a (along the incline)= gsin(10) = 10sin(10) = 1.74
v^2 = u^2 + 2as
15^2 = 3^2 + 2(1.74)s
s = 62.06 m
(way 2) using conservation of energy
energy (KE+PE) on top = energy at bottom
0.5m3^2 + mgh = 0.5m15^2 +0
h (height of incline) = (112.5 - 4.5)/10 = 19.8 m
length of incline = h/sin(10) = 62.2 m ; trigonometry
... find time
s = (u+v)t/2
t = 2s/(u+v) = 2(62.2)/(3+15) = 6.91 s
Answer:
Hi Carter,
The complete answer along with the explanation is shown below.
I hope it will clear your query
Pls rate me brainliest bro
Explanation:
The magnitude of the magnetic field on the axis of a circular loop, a distance z from the loop center, is given by Eq.:
B
= NμοiR² / 2(R²+Z²)³÷²
where
R is the radius of the loop
N is the number of turns
i is the current.
Both of the loops in the problem have the same radius, the same number of turns, and carry the same current. The currents are in the same sense, and the fields they produce are in the same direction in the region between them. We place the origin at the center of the left-hand loop and let x be the coordinate of a point on the axis between the loops. To calculate the field of the left-hand loop, we set z = x in the equation above. The chosen point on the axis is a distance s – x from the center of the right-hand loop. To calculate the field it produces, we put z = s – x in the equation above. The total field at the point is therefore
B
= NμοiR²/2 [1/ 2(R²+x²)³÷² + 1/ 2(R²+x²-2sx+s²)³÷²]
Its derivative with respect to x is
dB
/dx= - NμοiR²/2 [3x/ (R²+x²)⁵÷² + 3(x-s)/(R²+x²-2sx+s²)⁵÷²
]
When this is evaluated for x = s/2 (the midpoint between the loops) the result is
dB
/dx= - NμοiR²/2 [3(s/2)/ (R²+s²/4)⁵÷² - 3(s/2)/(R²+s²/4)⁵÷²
] =0
independent of the value of s.
None of the tables that you have attached is a correct one.
