Answer:
height is 69.68 m
Explanation:
given data
before it hits the ground = 46 % of entire distance
to find out
the height
solution
we know here acceleration and displacement that is
d = (0.5)gt² ..............1
here d is distance and g is acceleration and t is time
so when object falling it will be
h = 4.9 t² ....................2
and in 1st part of question
we have (100% - 46% ) = 54 %
so falling objects will be there
0.54 h = 4.9 (t-1)² ...................3
so
now we have 2 equation with unknown
we equate both equation
1st equation already solve for h
substitute h in the second equation and find t
0.54 × 4.9 t² = 4.9 (t-1)²
t = 0.576 s and 3.771 s
we use here 3.771 s because 0.576 s is useless displacement in the last second before it hits the ground is 46 % of the entire distance it falls
so take t = 3.771 s
then h from equation 2
h = 4.9 t²
h = 4.9 (3.771)²
h = 69.68 m
so height is 69.68 m
The type of figurative language that Freneau employ in these lines from "The Wild Honeysuckle" is personification. The correct answer would be option D. Why is it personification? Personification is a figure of speech that uses human attributes to something that is not living. In this line, the author attributes the words "roving" and "busy" to foot and hand, respectively.
The volume of the room is the product of its dimensions:

Now, from the equation

where d is the density, m is the mass and V is the volume, we deduce

So, multiply the density and the volume to get the mass of air in the room.
The acceleration produced in a body is always in the direction of the resultant force acting on the body. Therefore, we may determine the horizontal acceleration using the horizontal force applied. To do this, we may apply the mathematical form of Newton's second law:
Force = mass * acceleration
acceleration = force / mass
Substituting the values,
a = 100 / 0.15
a = 666.7 m/s²
The acceleration of the hockey puck is 670 m/s²
Answer:

Explanation:
We are given that
Frequency,f=800KHz=

Distance,d=4.5 km=
1 km=1000 m
Electric field,E=0.63V/m
We have to find the magnetic field amplitude of the signal at that point.

We know that


