Answer:
Watermelon is going at a speed of 
in downward direction
Explanation:
Given
- Height of the Empire State building = 320 m
- Speed of the superman = 24 m/s
since both the motion of both watermelon and the superman is under gravity. The acceleration of both is g in downwards direction. So the relative acceleration between them is zero and when they meet the relative distance covered by them will be the height of the Empire State building.
so let t be the time when they meet given by
Now let v be the velocity of the the watermelon when they meet given by
Answer:
The cross-sectional area of the larger piston is 392 cm²
Explanation:
Given;
output mass of the piston, m₀ = 2000 kg
input force of the piston, F₁ = 500 N
input area of the piston, A₁ = 10 cm² = 0.001 m²
The output force is given by;
F₀ = m₀g
F₀ = 2000 x 9.8
F₀ = 19600 N
The cross-sectional area of the larger piston or output area of the piston will be calculated by applying the following equations;
Therefore, the cross-sectional area of the larger piston is 392 cm²
Answer:
897
Explanation:
Speed of the car, v = 126 km/h, converting to m/s, we have v = 35 m/s and
Radius of the curve, R = 150 mm = 0.15 m
The centripetal acceleration a(c) is given by the formula = v² / R so that
a(c) = 35² / 0.15
a(c) = 1225 / 0.15
a(c) = 8167 m/s²
The force that causes the acceleration is frictional force = µ m g, where
µ = coefficient of friction
m = the mass of the car and
g = acceleration due to gravity, 9.81
From Newton's law:
µ m g = m a(c) , so that
µ = a(c) / g
µ = 8167 / 9.81
µ = 897
Therefore, the coefficient of static friction must be as big as 897
Answer:
a) I = (
+ 
) L² , b) w = (\frac{27 M}{18 m} + 2)⁻¹ Lv₀
Explanation:
a) The moment of inertia is a scalar that represents the inertia in circular motion, therefore it is an additive quantity.
The moment of inertia of a rod held at one end is
I₁ = 1/3 M L²
The moment of inertia of the mass at y = L
I₂ = m y²
The total inertia method
I = I₁ + I₂
I = \frac{1}{3} M L² + m (\frac{2}{3} L)²
I = ( + ) L²
b) The conservation of angular momentum, where the system is formed by the masses and the bar, in such a way that all the forces during the collision are internal.
Initial instant. Before the crash
L₀ = I₂ w₀
angular and linear velocity are related
w₀ = y v₀
w₀ = 
L v₀
L₀ = I₂ y v₀
Final moment. After the crash
= I w
how angular momentum is conserved
L₀ = L_{f}
I₂ y v₀ = I w
substitute
m (
)² (\frac{2L}{3} v₀ = ( + ) L² w
m L³ v₀ = ( + ) L² w
m L v₀ = ( + ) w
L v₀ = 
w
w = (\frac{27 M}{18 m} + 2)⁻¹ Lv₀
