The speed of the toy when it hits the ground is 2.97 m/s.
The given parameters;
The speed of the toy before it hits the ground will be maximum. Apply the principle of conservation of mechanical energy to determine the maximum speed of the toy.
P.E = K.E
Substitute the given values and solve the speed;
Thus, the speed of the toy when it hits the ground is 2.97 m/s.
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Answer:
51.94 ft/s²
257.63 ft/s
Explanation:
t = Time taken = 4 s
u = Initial velocity = 34 mi/h
v = Final velocity
s = Displacement = 615 ft
a = Acceleration
Converting velocity to ft/s
Equation of motion
Acceleration is 51.94 ft/s²
Final velocity at this time is 257.63 ft/s
Answer:
V_{a} - V_{b} = 89.3
Explanation:
The electric potential is defined by
= - ∫ E .ds
In this case the electric field is in the direction and the points (ds) are also in the direction and therefore the angle is zero and the scalar product is reduced to the algebraic product.
V_{b} - V_{a} = - ∫ E ds
We substitute
V_{b} - V_{a} = - ∫ (α + β/ y²) dy
We integrate
V_{b} - V_{a} = - α y + β / y
We evaluate between the lower limit A 2 cm = 0.02 m and the upper limit B 3 cm = 0.03 m
V_{b} - V_{a} = - α (0.03 - 0.02) + β (1 / 0.03 - 1 / 0.02)
V_{b} - V_{a} = - 600 0.01 + 5 (-16.67) = -6 - 83.33
V_{b} - V_{a} = - 89.3 V
As they ask us the reverse case
V_{b} - V_{a} = - V_{b} - V_{a}
V_{a} - V_{b} = 89.3