Answer:
The ball's initial kinetic energy
The ball comes to a stop at B. At this point its initial kinetic energy is converted into potential energy
Explanation:
A ball is fixed to the end of a string, which is attached to the ceiling at point P. As the drawing shows, the ball is projected downward at A with the launch speed v0. Traveling on a circular path, the ball comes to a halt at point B. What enables the ball to reach point B, which is above point A? Ignore friction and air resistance.
From conservation of energy which states that energy can neither be created nor be destroyed, but can be transformed from one form to another.
Ki+Ui=Kf+Uf
Ki=initial kinetic energy
Ui=initial potential energy
Kf=final kinetic energy
Uf=final potential energy
we know that
m=mass of the ball
ha=downward height a
hb=upward height b
u=initial velocity u
v=final velocity v, which is 0
g=acceleration due to gravity
v=0 at final velocity
1/2mu^2+mgha=0+1/2mv^2
ha=hb+Ki/mh
From the above equation, we can conclude that the ball's initial kinetic energy is responsible for making the ball reach point B.
Point B is higher than point A from the motion gained by the ball
Answer:
shock, tremble, fault, slip, temblor, upheaval, trembler
undulation
macroseism
seimicity
seismism
Explanation:
Answer:
total stretch of the double-length spring will be 20 cm
Explanation:
given data
length x1 = 10 cm
mass = 1 kg
mass = double = 2 kg
to find out
the total stretch of the double-length spring will be
solution
we can say here spring constant is
k = mg ............1
k is spring constant and m is mass and g is acceleration due to gravity
so for in 1st case and 2nd case with 1 kg mass and 2 kg mass
kx1 = mg .........................2
and
kx2 = 2mg ........................3
x is length
so from equation 2 and 3
x2 = 20
so total stretch of the double-length spring will be 20 cm
Answer:
a) 10.54 sec
b) 284.58 m
c) 29.406 m/s
d) 39.92 m/s
Explanation:
Given data:
velocity of spacecraft = 27.0 m/s
rate of free fall acceleration is 2.79 m/s^2
distance of moving aircraft from mooon surface is 155 m
a. from kinematic eqaution of motion we have
where y = 155 m
Vi = 0 as this relation is for vertical motion, so the 27.0 m/s is not included
and a = 2.79 m/s^2.
Solving for t we get
t = 10.54 sec
b.
we know that
c. from the kinematic formula
v = u + at
v = 29.4066 m/a
d.
v = 39.92 m/s