Explanation:
A boat heads north in still water at 4.5 m/s directly across a river that is running east at 3.0 m/s.
Let north be positive y axis and east be positive x axis.
We have
Velocity of boat = 4.5 j m/s
Velocity of river = 3 i m/s
Velocity of the boat with respect to Earth = 3 i + 4.5 j
That is 56.31° north of east.
Velocity of the boat with respect to Earth is 5 m/s 56.31° north of east
Answer:
5.41 m/s
Explanation:
velocity of boat with respect to water , Vb,w = 4.5 j
velocity of water with respect to ground, Vw,g = 3 i
velocity of boat with respect to water = velocity of boat - velocity of water
Vb,w = Vb, g - Vw, g
4.5 i = Vb,g - 3i
Vb,g = 3 i + 4.5 j
The magnitude of velocity = = 5.41 m/s
Answer:
Please refer to the figure.
Explanation:
The magnitude of the magnetic field can be found by Biot-Savart Law. We should divide the loop into four components. Each component has a similar solution but their directions are quite different.
The directions can be found by right-hand rule. Point your index finger into the direction of current, point your middle finger towards the target point (0,0,a). Your thumb will show you the direction of magnetic field.
Answer:
a) the total electric potential is 2282000 V
b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
Explanation:
Given the data in the question and as illustrated in the image below;
a) Determine the total electric potential (in V) at the origin.
We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges
so
Electric potential at p in the diagram 1 below is;
Vp = V1 + V2
Vp = kq1/r1 + kq2/r2
we know that; Coulomb constant, k = 9 × 10⁹ C
q1 = 4.60 uC = 4.60 × 10⁻⁶ C
r1 = 1.25 cm = 0.0125 m
q2 = -2.06 uC = -2.06 × 10⁻⁶ C
location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m
so we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )
Vp = (3312000) + ( -1030000 )
Vp = 3312000 -1030000
Vp = 2282000 V
Therefore, the total electric potential is 2282000 V
b)
the total electric potential (in V) at the point with coordinates (0, 1.50 cm).
As illustrated in the second image;
r1² = 0.015² + 0.0125²
r1 = √[ 0.015² + 0.0125² ]
r1 = √0.00038125
r1 = 0.0195
Also
r2² = 0.015² + 0.018²
r2 = √[ 0.015² + 0.018² ]
r2 = √0.000549
r2 = 0.0234
Now, Electric Potential at P in the second image below will be;
Vp = V1 + V2
Vp = kq1/r1 + kq2/r2
we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )
Vp = 2123076.923 + ( -762962.962 )
Vp = 2123076.923 -792307.692
Vp = 1330769.23 V
Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
