Am sorry what can you be more specific
Answer:
W=1055N
Explanation:
In order to solve this problem, we must first do a drawing of the situation so we can visualize theh problem better. (See attached picture)
In this problem, we will ignore the board's weight. As we can see in the free body diagram of the board, there are only three forces acting on the system and we can say the system is in vertical equilibrium, so from this we can say that:
so we can do the sum now:
when solving for the Weight W, we get:
and now we can substitute the given data, so we get:
W=410N+645N
W=1055N
Answer:
Magnitude of the force on proton = F = 1.1085 × 10^-15 N
Explanation:
Charge on proton = q = 1.60 × 10^-19 C
Velocity of proton = V = 4.0 × 10^4 m/s
Magnetic field = B = 0.20 T
Angle between V and B = θ = 60
We know that,
F = qVBsin θ = (1.60 × 10^-19)( 4.0 × 10^4)( 0.20)sin(60)
F = 1.1085 × 10^-15 N
Responder:
A. Ff = 300 N N = 784,8 N
Explicación:
Dado
Masa del cuerpo = 80 kg
Fuerza de movimiento Fm = 300N
Dado que el cuerpo no está acelerando, la fuerza de fricción (Ff) es igual a la fuerza de movimiento que actúa sobre el cuerpo, ya que la fuerza de fricción es una fuerza de oposición, es decir, Fm = Ff
Dado que Fm = 300N, Ff = 300N
La reacción normal que actúa en el cuerpo es igual al peso.
N = W = mg
g es la aceleración debida a la gravedad
g = 9,8 m / s
N = mg
N = 80 (9,81)
N = 784,8N
Por tanto, la fuerza normal que actúa sobre el cuerpo es 784,8 N
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