Answer:0.4 m
Explanation:
Given
Maximum displacement A=0.49
The sum of kinetic and elastic potential energy is 
where k=spring constant
U+K.E.=
when K.E.=U/2
K.E.=kinetic energy
U=Elastic potential Energy

Answer:
0.8 x 10^-9 kg
Explanation:
Given,
Distance ( R ) = 10 m
Force ( F ) = 3.2 x 10^-9 N
Mass ( m1 ) = 40 kg
To find : Mass ( m2 ) = ?
Formula : -
F = m1.m2 / R^2
m2 = FR^2 / m1
= 3.2 x 10^-9 x 10 / 40
= 3.2 x 10^-9 / 4
= ( 3.2 / 4 ) x 10^-9
m2 = 0.8 x 10^-9 kg
Answer:

Explanation:
information we know:
Total force: 
Weight: 
distance: 
vertical component of the force: 
-------------
In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).
horizontal component: 
vertical component: 
but from the given information we know that 
so, equation these two
and 

and we know the force
, thus:

now we clear for 

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:


whith this horizontal component we calculate the work to move the crate a distance of 4 m:

the work done is W=173.48J
What are the choices ?
Without some directed choices, I'm, free to make up any
reasonable statement that could be said about Kevin in this
situation. A few of them might be . . .
-- Kevin will have no trouble getting back in time for dinner.
-- Kevin will have no time to enjoy the scenery along the way.
-- Some simple Physics shows us that Kevin is out of his mind.
He can't really do that.
-- Speed = (distance covered) / (time to cover the distance) .
If time to cover the distance is zero, then speed is huge (infinite).
-- Kinetic energy = (1/2) (mass) (speed)² .
If speed is huge (infinite), then kinetic energy is huge squared (even more).
There is not enough energy in the galaxy to push Kevin to that kind of speed.
-- Mass = (Kevin's rest-mass) / √(1 - v²/c²)
-- As soon as Kevin reaches light-speed, his mass becomes infinite.
-- It takes an infinite amount of energy to push him any faster.
-- If he succeeds somehow, his mass becomes imaginary.
-- At that point, he might as well turn around and go home ...
if he ever reached Planet-Y, nobody could see him anyway.
An object distance is
presented as s = 5f and we know that the mirror equation relates the image
distance to the object distance and the focal length.
The mirror equation is
1/f = 1/s + 1/s’ where the variable f stands for
the focal length of the mirror. Variable (s)
represents the distance between the mirror surface and the object and the
variable <span>(s’) represents the distance between the mirror surface and
the image. </span>
In addition, a concave mirror
will have a positive focal length (f) and a convex mirror will have a negative
focal length (f).
Now, we then have 1/f = 1/5f
+ 1/s’ which is s’ = 5f/4
Then we get the magnification
ratio that expresses the size or amount of magnification or reduction of the
object or image and to get the magnification, we use this equation: M= s’/s
M= 5f/4x5f
s’ = 1/4s
Therefore, the image height
is one fourth of the object height