Question

Шарик массой 0,1 кг,подвешенный на нити 0,5м,равномерно вращается в вертикальной плоскости со скоростью 3 м/с. Каков вес шарика в нижней точке траектории?

Answer 1

P=m*(g+a)=m*(g+V^2/L)=0,1*(10+9/0,5)=2,8 H

Answer 2

$P = mg + \frac{mv^2}{R}$

m - масса ( 0,1 кг )
g - ускорение свободного падения ( 10 Н/кг )
v - скорость ( 3 м/c )
R - радиус ( 0,5 м )

$P = 0,1*10 + \frac{0,1*3^2}{0,5} = 1 + \frac{0,1*9}{0,5} = 1 + \frac{0,9}{0,5}$ = 1 + 1,8 = 2,8 H 

Answer 3

Calculation of the Weight of the Ball at the Bottom of its Trajectory

To calculate the weight of the ball at the bottom of its trajectory, we need to consider the forces acting on the ball. At the bottom of the trajectory, the tension in the string provides the centripetal force required to keep the ball moving in a circular path.

The centripetal force is given by the equation:

F = m * v^2 / r

Where: - F is the centripetal force - m is the mass of the ball - v is the velocity of the ball - r is the radius of the circular path

In this case, the radius of the circular path is equal to the length of the string, which is 0.5 meters.

Given that the mass of the ball is 0.1 kg and the velocity is 3 m/s, we can calculate the centripetal force:

F = (0.1 kg) * (3 m/s)^2 / 0.5 m

Simplifying the equation:

F = 0.1 kg * 9 m^2/s^2 / 0.5 m

F = 1.8 N

Therefore, the weight of the ball at the bottom of its trajectory is 1.8 Newtons.