Answer:
v = 17.71 m / s
Explanation:
We can work this exercise with the kinematics equations. In general the body is released so that its initial velocity is zero, the acceleration of the acceleration of gravity
v² = v₀² - 2 g (y -y₀)
v² = 0 - 2g (y -y₀)
when it hits the stone the height is zero and part of the height of the seagull I
v² = 2g y₀
v = Ra (2g i)
let's calculate
v =√ (2 9.8 16)
v = 17.71 m / s
Answer:
The frequency of the oscillation is 0.9Hz
Explanation:
This problem bothers on simple harmonic motion of a spring
Given data
Mass of the child m= 25kg
Spring constant k=791 N/m
Amplitude a= 31cm
But the period of the motion as a result of the adults sholve is expressed as
T=2π√m/k
T=2*3.142√25/791
T=6.284√0.031
T=6.284*0.176
T=1.11 sec
But frequency F=1/T
F=1/1.11
F=0.9Hz
As its charge, proton -a positive charged molecule at the center of an atom- is the opposite of the electron -the particle which is orbiting the center of an atom.
Answer:
the final velocity of the car is 59.33 m/s [N]
Explanation:
Given;
acceleration of the car, a = 13 m/s²
initial velocity of the car, u = 120 km/h = 33.33 m/s
duration of the car motion, t = 2 s
The final velocity of the car in the same direction is calculated as follows;
v = u + at
where;
v is the final velocity of the car
v = 33.33 + (13 x 2)
v = 59.33 m/s [N]
Therefore, the final velocity of the car is 59.33 m/s [N]
