The net force acting on the crate is determined as 176 N to the left.
<h3>Net force acting on the crate</h3>
The net force acting on the crate is calculated as follows;
∑F = F1 + F2 + F3 + F4
F(net) = -440y + 176x + 440y - 352x
F(net) = -176 x
The resultant force is pointing in negative x direction.
Thus, the net force acting on the crate is determined as 176 N to the left.
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<span>Mass of the ball is m = 0.10kg
Initial speed of the Ball v = 15m/s
a. When the ball is at maximum height the velocity is 0
Momentum of ball = mass x velocity
Momentum = 0.10kg x 0 = 0
b. Getting the maximum height,
Using the conservation of energy equation KEinitial = mgh
1/2mVin^2 = mgh => h = v^2/2g
h = 15^2/2x9.8 = 11.48m => Half Height h = 5.96m
Applying the conservation of energy equation at halfway V^2 = 2gh
V = square root of (2x9.8x5.96) => V = square root of (116.816)
So the velocity at the half way V = 10.81 m/s
Momentum M = m x V => M = 0.10 x 10.81 => M = 1.081kg-m/s</span>
Answer:
i) 24.5 m/s
ii) 30,656 m
iii) 89,344 m
Explanation:
Desde una altura de 120 m se deja caer un cuerpo. Calcule a 2.5 s i) la velocidad que toma; ii) cuánto ha disminuido; iii) cuánto queda por hacer
i) Los parámetros dados son;
Altura inicial, s = 120 m
El tiempo en caída libre = 2.5 s
De la ecuación de caída libre, tenemos;
v = u + gt
Dónde:
u = Velocidad inicial = 0 m / s
g = Aceleración debida a la gravedad = 9.81 m / s²
t = Tiempo de caída libre = 2.5 s
Por lo tanto;
v = 0 + 9.8 × 2.5 = 24.5 m / s
ii) El nivel que el cuerpo ha alcanzado en 2.5 segundos está dado por la relación
s = u · t + 1/2 · g · t²
= 0 × 2.5 + 1/2 × 9.81 × 2.5² = 30.656 m
iii) La altura restante = 120 - 30.656 = 89.344 m.
Answer:
a single closed path of electrical components including a voltage source
We can solve this using Snell's Law which is represented by the equation:
sin θ₁ / sin θ₂ = n₂ / n₁
From the problem, we can substitute values and solve for the angle of refraction.
sin 19 / sin θ₂ = 1.65 / 1
θ₂ = 11.38°
The angle of refraction would be 11.38°.
