Answer:
2,250J
Explanation:
W = Fs = (450)(5) = 2,250
Answer:
The induced current is 26.7 mA
Explanation:
Given;
area of the loop, A = 0.078 m²
initial magnetic field, B₁ = 3.8 T
change in the magnetic field strength, dB/dt = 0.24 T/s
The induced emf is calculated as;

The resistance of the loop = 0.7 Ω
The induced current is calculated as;

Answer:
1. the electric potential energy of the electron when it is at the midpoint is - 2.9 x
J
2. the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge is - 5.04 x
J
Explanation:
given information:
= 3 nC = 3 x
C
= 2 nC = 2 x
C
r = 50 cm = 0.5 m
the electric potential energy of the electron when it is at the midpoint
potential energy of the charge, F
F = k 
where
k = constant (8.99 x
)
electron charge,
= - 1.6 x
C
since it is measured at the midpoint,
r = 
= 0.25 m
thus,
F = 
= k
+ k
=
(
)
= (8.99 x
)( - 1.6 x
)(3 x
+2 x
)/0.25
= - 2.9 x
J
the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge
= 10 cm = 0.1 m
= 0.5 - 0.1 = 0.4 m
F = k
+ k
=
(
+
)
= (8.99 x
)( - 1.6 x
)(3 x
/0.1+2 x
/0.4)
= - 5.04 x
J
Answer:
A) 35 ft
B) 5 ft
C) Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick
Explanation:
A) Total distance covered by the dog = 20 + 15
= 35 ft
B) Since the other distance covered by the dog before chewing the stick, after the retrieval, was in an opposite direction to the initial direction, then;
total displacement of the dog = 20 - 15
= 5 ft
C) Net displacement = distance covered by the dog to retrieve the stick + distance covered before the dog starts chewing the stick
But, displacement involves a specified direction. The distance covered before the dog starts chewing the stick was in an opposite direction to the initial direction.
Thus,
Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick