Answer:
The smallest possibility is 0.01E-22kgm/s
Explanation:
Using
Momentum= h/4πx
= 6.6x 10^-34Js/ 4(3.142* 50*10-12m)
= 0.01*10^-22kgm/s
It is called a photon i believe
It accelerates in the y component (bc of gravity) AND the x-component (b/c of the friction force).
Answer:
y₀ = 10.625 m
Explanation:
For this exercise we will use the kinematic relations, where the upward direction is positive.
y = y₀ + v₀ t - ½ g t²
in the exercise they indicate the initial velocity v₀ = 8 m / s.
when the rock reaches the ground its height is zero
0 = y₀ + v₀ t - ½ g t²
y₀i = -v₀ t + ½ g t²
let's calculate
y₀ = - 8 2.5 + ½ 9.8 2.5²
y₀ = 10.625 m
when the apple moves in a horizontal circle, the tension force in the string provides the necessary centripetal force to move in circle. the tension in the string is given as
T=mv²/r
where T = tension force in the string , m = mass of the apple
v = speed of apple , r = radius of circle.
clearly , tension force depends on the square of the speed. hence greater the speed, greater will be the tension force.
at some point , the speed becomes large enough that it makes the tension force in the string becomes greater than the tensile strength of the string. at that point , the string breaks
