the friction forces are smaller than the forward force
The kayaker has velocity vector
<em>v</em> = (2.50 m/s) (cos(45º) <em>i</em> + sin(45º) <em>j</em> )
<em>v</em> ≈ (1.77 m/s) (<em>i</em> + <em>j</em> )
and the current has velocity vector
<em>w</em> = (1.25 m/s) (cos(315º) <em>i</em> + sin(315º) <em>j</em> )
<em>w</em> ≈ (0.884 m/s) (<em>i</em> - <em>j</em> )
The kayaker's total velocity is the sum of these:
<em>v</em> + <em>w</em> ≈ (2.65 m/s) <em>i</em> + (0.884 m/s) <em>j</em>
That is, the kayaker has a velocity of about ||<em>v</em> + <em>w</em>|| ≈ 2.80 m/s in a direction <em>θ</em> such that
tan(<em>θ</em>) = (0.884 m/s) / (2.65 m/s) → <em>θ</em> ≈ 18.4º
or about 18.4º north of east.
The magnetic part using the Lorentz force is: F = q v x
B,
where v and B are vectors and v x B is the vector cross product.
Magnitude of the force: F = q v B sin(α)
So, sin(α) = F/( e v B), with e the proton charge.
This will give you a value for sin(α), and two potentials
for its opposite.
You will now look for:
sin(α) = 7.40 10^-13/( 1.60 10^-19 * 5 10^6 * 1.78)
= 0.520
So either sin(α) = 0.502 or sin(α) = -0.502
The 1st α = 30.1 degrees or α = 150 degrees.
The 2nd α = 210 degrees or α = 330 degrees.
So we can say that 30.1 degrees and 330 degrees would be minimum and biggest on [0,360]
Answer:
46.8 kg
Explanation:
Mass = (density)(volume)
= (1.3)(36)
<u>M</u><u>a</u><u>s</u><u>s</u><u> </u><u>=</u><u> </u><u>4</u><u>6</u><u>.</u><u>8</u><u> </u><u>k</u><u>g</u>
