<span>Example Problems. Kinetic Energy (KE = ½ m v2). 1) The velocity of a car is 65 m/s and its mass is 2515 kg. What is its KE? 2) If a 30 kg child were running at a rate of 9.9 m/s, what is his KE? Practice Problems. IN THIS ORDER…. Page 2: #s 6, 7, 8, 5. Potential Energy. An object can store energy as the result of its position.</span><span>
</span>
Answer:
Explanation:
side of the square loop, a = 7 cm
distance of the nearest side from long wire, r = 2 cm = 0.02 m
di/dt = 9 A/s
Integrate on both the sides
i = 9t
(a) The magnetic field due to the current carrying wire at a distance r is given by
(b)
Magnetic flux,
(c)
R = 3 ohm
magnitude of voltage is
e = 1.89 x 10^-7 V
induced current, i = e / R = (1.89 x 10^-7) / 3
i = 6.3 x 10^-8 A
Answer:
Explanation:
Using the pythagoras theorem, the displacement is expressed as;
d² = x²+y²
y = 36m (north)
x = 20m east
Substitute;
d² = 36²+20²
d² = 1296+400
d² = 1696
d = √1696
d = 41.18m
For the direction;
theta = tan^-1(y/x)
theta = tan^-1(36/20)
theta = tan^-1(1.8)
theta = 60.95°
Hence the magnitude is 41.18m and the direction is 60.95°
