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.
Answer:
Maximum height, h = 11.32 meters
Explanation:
It is given that,
The baseball is thrown directly upward at time, t = 0
Initial speed of the baseball, u = 14.9 m/s
Ignoring the resistance in this case and using a = g = 9.8 m/s²
We have to find the maximum height the ball reaches above where it leaves your hand. Let the maximum height is h. Using third equation of motion as :

At maximum height, v = 0
and a = -g = -9.8 m/s²


h = 11.32 meters
Hence, the maximum height of the baseball is 11.32 meters.
Answer:
Internal resistance = 0.545 ohm
Explanation:
As per ohm's law we know that

here we know that
i = electric current = 2.75 A
V = potential difference = 1.50 Volts
now from above equation we have

now we have


The kinetic energy of the object is 392 Joules.
It is more likely to be a metal. This is because metals are good heat conductors whereas non metals tend to be poor heat conductors.