it can be said that the speed of the east wind is
v=0.3608m/s
From the question we are told
A small boat sailed <u>straight </u>north out of a harbor in <em>strong </em>east wind (blowing from west to east).
After sailing for 120 minutes, it ended up hitting a buoy 60^\circ60 ∘ to the north-east of the harbor. If the straight-line distance between the buoy and the harbor is 3 km,
- what is the speed of the east wind?.
<h3> the speed of the east wind</h3>
Generally the equation for the distance is mathematically given as
BA=3000sin60
BA=2598.07m
Therefore
the speed of the east wind

v=0.3608
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Would be A 1012 N/C because The magnitude of the electric field at distance r from a point charge q is E=k
e
q/r
2
, so
E=
(5.11×10
−11
m)
2
(8.99×10
9
N.m
2
/C
2
)(1.60×10
−19
C)
=5.51×10
11
N/C∼10
1
2N/C
making (e) the best choice for this question.
Answer:
Wavelength of light is 600 nm
Explanation:
Given
Distance between the first order maximum and the tenth order maximum of a double-slit pattern = 18 mm
Separation between the slits = 0.15 mm
Distance of screen from the slits = 50 cm
Wavelength
nm
Answer:
I = 1.875 A
Explanation:
For this exercise we use Ampere's law
∫ B . ds = μ₀ I
We use a circular path around the wire whereby B and ds are parallel, whereby the dot product is reduced to the algebraic product
ds = 2π dr
B (2πr) = μ₀ I
I = B 2π R /μ₀
r= 7.5 cm = 0.075 m
calculate
I = (50 μ₀ /π) 2π 0.075 /μ₀
I = 1.875 A