Potential energy is energy stored in an object. kinetic energy is energy of motion
Answer:
Point 3
Explanation:
It is a summer month position but it has no sunlight.
Answer:
I = 0.5 A
Explanation:
Given: P=60 Watts, Voltage supply V = 120 Volts (for primary coil)
Solution:
we have P = V I
⇒ I = P /V = 60 Watts / 120 Volts
I = 0.5 A
Most waves approach the shore at an angle. However, they bend to be nearly parallel to the shore as they approach it because when a wave reaches a beach or coastline, it releases a burst of energy that generates a current, which runs parallel to the shoreline.
- Most waves approach shore at an angle. As each one arrives, it pushes water along the shore, creating what is known as a longshore current within the surf zone.
- Waves approach the coast at an angle because of the direction of prevailing wind.
- The part of the wave in shallow water slows down, while the part of the wave in deeper water moves at the same speed.
- Thus when wave reaches a beach or coastline, it releases a burst of energy that generates a current, which runs parallel to the shoreline.
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When we say directly northeast that is equivalent
to 45˚ north of east.
First let us determine the north and east
components of the acceleration using cos and sin functions:<span>
North = 2.18 * sin 45
East = 2.18 * cos 45
<span>Then we set to determine the east component of the plane’s
displacement by calculating using the formula:
d = vi * t + ½ * a * t^2
d = 135 * 18 + ½ * 2.18 * cos 45 * 18^2
<span>d = 2430 + 353.16 * cos 45 = 2679.72 m</span>
Calculating for the north component:
North = ½ * 2.18 * sin 45 * 18^2 </span></span>
North = 249.72 m
Hence magnitude is:
Magnitude = sqrt (2679.72^2 + 249.72^2)
Magnitude = 2,691. 33 m<span>
</span>
Calculating for angle:
Tan θ = North ÷ East <span>
<span>Tan θ = 249.72 m ÷ 2679.72 m</span></span>
θ = 5.32°<span>
</span>
So the plane was flying at 2,691. 33 m at 5.32<span>°</span>
