Answer:
<em>7.45 10^5N.</em>
Explanation:
according to newtons second law of motion;
Fapp is the applied force
Fr is the resistive force
m is the mass of the luxury
a is the acceleration
Since the huge luxury liner move with constant velocity, then acceleration is zero i.e a = 0. The equation becomes;
This shows that the applied force will be equal to the resistive force if the velocity is constant.
Given Fr = 7.45 10^5 N therefore the resistive force will also be 7.45 10^5N.
<em>Hence the magnitude of the resistive force exerted by the water on the cruise ship is 7.45 10^5N.</em>
Answer:
7.85 m/s^2
Explanation:
linear or tangential acceleration= dv/dt
⇒
=0.83 m/s^2
radial acceleration is given by =
⇒
= 7.81 m/s^2
total acceleration
putting values we get
= 7.85 m/s^2
Answer:
Super idoo di shaw lung domini di shaw
Answer:
Eje x
fr- F = m a
eje y
N-W =0
Explanation:
Para este ejercicio, debemos usar la segunda ley de Newton, donde se necesita fija un sistema de referencia el mas usado es un sistema horizontal y vertical para los ejes x e y
en el adjunto puede ver el diagrama de cuerpo libre correcto,
las fuerza son
Eje x
fr- F = m a
eje y
N-W =0
la ecuacion para la fuerza de roce es
fr = my N