Answer:
2.43J
Explanation:
Given parameters:
Mass of the arrow = 0.155kg
Velocity = 31.4m /s
Unknown:
Kinetic energy when it leaves the bow = ?
Solution:
The kinetic energy of a body is the energy in motion of the body;
it can be derived using the expression below:
K.E = m v²
m is the mass
v is the velocity
Solve for K.E;
K.E = x 0.155 x 31.4 = 2.43J
Answer:
Explanation:
From the given information:
The coordinate axis is situated in the east and north direction.
So, the north will be the y-axis and the east will be the x-axis
Similarly, the velocity of the plane in regard to the air in the coordinate system will be
where:
= velocity of the plane in regard to the air
v = velocity
θ = angle of inclination of the plane with respect to the horizontal
replacing v = 180 km/ and θ = 20° in above equation, then:
The velocity of the airplane in the coordinate system as:
where;
= velocity of the airplane
= velocity
∅ = angle of inclination with regard to the base axis;
Then; replacing = 150 km/h and ∅ = 30°
Therefore, the velocity of the plane in the system is :
--- (1)
( 150 cos 30° - 180 cos 20°)i + ( 150 sin 30° - 180sin 20°)j
(-39.24 km/h)i + (13.44 km/h) j
The magnitude is:
= 41.48 km/h
The airplane is moving at an angle of the inverse tangent to the abscissa and ordinate.
The angle of motion is:
tan θ = 39.24/13.44
tan θ = 2.9
θ =
θ = 70.97°
The angle of motion is 70.97° from west of north with a velocity of 41.48 km/h.