Let <em>a</em> denote the airplane's velocity in the air, <em>g</em> its velocity on the ground, and <em>w</em> the velocity of the wind. (Note that these are vectors.) Then
<em>a</em> = <em>g</em> + <em>w</em>
and we're given
<em>a</em> = (325 m/s) <em>j</em>
<em>w</em> = (55.0 m/s) <em>i</em>
Then
<em>g</em> = - (55.0 m/s) <em>i</em> + (325 m/s) <em>j</em>
The ground speed is the magnitude of this vector:
||<em>g</em>|| = √[ (-55.0 m/s)² + (325 m/s)² ] ≈ 330. m/s
which is faster than the air speed, which is ||<em>a</em>|| = 325 m/s.
Answer: Option (b) is correct.
Explanation:
Since we know that,
P = VI
where;
P = power
V= Voltage
I = Current
Since it's given that,
P = 600W
I = 2.5 A
equating these values in the above equation, we get;
<em>V = 
</em>
<em>V = 240 V</em>
paraan ito ng pagrerespeto
If the object's kinetic energy is zero, then due to in multiplication factor, it's momentum will also be equal to zero 'cause the velocity of the object must be Nil
In short, Your Answer would be: "Zero"
Hope this helps!
