Answer:
v = 9.07 m/s
the vertical component of the velocity of the fish relative to the water when it hits the water is 9.07 m/s
Explanation:
Given;
An eagle is flying horizontally at a speed of 4.6 m/s
Initial horizontal velocity uh = 4.6 m/s
Initial vertical velocity uy = 0
Height to fall d = 4.2 m
Acceleration due to gravity g = 9.8 m/s^2
The final vertical velocity of the fish when it hits the water can be calculated using the equation of motion;
v^2 = u^2 + 2as
v^2 = uy^2 + 2gd
uy = 0
v^2 = 2gd
v = √(2gd)
Substituting the given values;
v = √(2×9.8×4.2)
v = 9.073036977771 m/s
v = 9.07 m/s
the vertical component of the velocity of the fish relative to the water when it hits the water is 9.07 m/s
Answer: M^-1 L^-3T^4A^2
Explanation:
From coloumb's law
K = q1q2 / (F × r^2)
Where;
q1, q2 = charges
k = constant (permittivity of free space)
r = distance
Charge (q) = current(A) × time(T) = TA
THEREFORE,
q1q2 = (TA) × (TA) = (TA)^2
Velocity = Distance(L) / time(T) = L/T
Acceleration = change in Velocity(L/T) / time (T)
Therefore, acceleration = LT^-2
Force(F) = Mass(M) × acceleration (LT^-2)
Force(F) = MLT^-2
Distance(r^2) = L^2
From ; K = q1q2 / (F × r^2)
K = (TA)^2 / (MLT^-2) (L^2)
K = T^2A^2M^-1L^-1T^2 L^-2
COLLEXTING LIKE TERMS
T^2+2 A^2 M^-1 L^-1-2
M^-1 L^-3T^4A^2
