Answer:
10.1 N
Explanation:
Your answer is 10.1 N, I don't actually know how to do it but I hope it helps.
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Answer:
El peso del cartel es 397,97 N
Explanation:
La tensión dada en cada segmento del cable = 2000 N
El desplazamiento vertical del cable = 50 cm = 0,5 m
La distancia entre los polos = 10 m
La posición del letrero en el cable = En el medio = 5
El ángulo de inclinación del cable a la vertical = tan⁻¹ (0.5 / 5) = 5.71 °
El peso del letrero = La suma del componente vertical de la tensión en cada lado del letrero
El peso del signo = 2000 × sin (5.71 grados) + 2000 × sin (5.71 grados) = 397.97 N
El peso del signo = 397,97 N.
The situation given above can be answered through the concept of the First Law of thermodynamics which states that the change in internal energy is equal to the difference between the work done and the heat added to the system. The work done by the object is negative and the heat added is positive.
change in internal energy = -500J + 1400 J = 900 J
Answer:
The the maximum force acting on the crate is 533.12 newtons.
Explanation:
It is given that,
Mass of the wooden crate, m = 136 kg
The coefficient of static friction, 
The coefficient of kinetic friction, 
We need to find the maximum force exerted horizontally on the crate without moving it. As the crate is not moving than the coefficient of static friction will act and the force is given by :
F = 533.12 N
So, the maximum force acting on the crate is 533.12 newtons. Hence, this is the required solution.
