Answer:
solution given:
acceleration (a)=?
initial velocity (u)=3m/s
final velocity (v)=6m/s
distance (s)=90m
we have
v²=u²+2as
substituting value
6²=3²+2*a*90
36=9+180a
36-9=180a
a=25/180
<u>a=0.1388m/s²</u>
The circumference of a circle is (2π · the circle's radius).
The length of a semi-circle is (1π · the circle's radius) =
(π · 14.8) = 46.5 (rounded)
(The unit is the same as whatever the unit of the 14.8 is.)
0.77 m/s2 directed 35° south of west
net force = (-17,-12)
net force = mass * acceleration
(-17,-12) = 27 * (x-acceleration,y-acceleration)
(x-acceleration,y-acceleration) = (-17/27,-12/27) = (-0.629629629..., -0.444...)
angle of acceleration = tan^-1 (-0.444.../-0.629629...) = 35.21759 degrees below negative x-axis.
magnitude of acceleration = sqrt((-0.629629...)^2 + (-0.444...)^2) = 0.77069 (5dp)
Answer: a) 11.76 m/s b) 7.056 m
Explanation:
The described situation is as follows:
An object is dropped from the top of a tower and when measuring the time it takes to reach the ground that turns out to be 0.02 minutes.
This situation is related to free fall, this also means we have constant acceleration, hence the equations we will use are:
(1)
(2)
Where:
Is the final velocity of the object
Is the initial velocity of the object (it was dropped)
is the acceleration due gravity
is the height of the tower
is the time it takes to the object to reach the ground
b) Begining with (1):
(3)
(4)
(5) This is the final velocity of the object
a) Substituting (5) in (2):
(6)
Clearing :
(7)
(8) This is the height of the tower
Electrons move in atomic orbitals (or subshells). there are four different orbital shapes (s p d f). in each shell, the s subshell is at a lower energy than the p. an orbital diagram is used to determine an atom's electron configurations
