Answer:
louder sound.
Explanation:
The amplitude of a sound wave determines its loudness or volume. A larger amplitude means a louder sound
Answer:
12.7m/s
Explanation:
Given parameters:
Mass of the diver = 77kg
Height = 8.18m
Unknown:
Final velocity = ?
Solution:
To solve this problem, we use one of the motion equations.
v² = u² + 2gh
v is the final velocity
u is the initial velocity
g is the acceleration due to gravity
h is the height
v² = 0² + (2 x 9.8 x 8.18)
v² = 160.3
v = 12.7m/s
To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.
Gravitational potential energy can be defined as

As M=m, then

Where,
m = Mass
G =Gravitational Universal Constant
R = Distance /Radius
PART A) As half its initial value is u'=2u, then



Therefore replacing we have that,

Re-arrange to find v,



Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s
PART B) With a final separation distance of 2r, we have that

Therefore




Therefore the velocity when they are about to collide is 
Answer:
28.15
Explanation:
the ateps is in the photo
Answer:
t = 3/8 seconds
Explanation:
h=-16t^2 - 10t+6
h= 0 when it hits the ground
0=-16t^2 - 10t+6
factor out a -2
0= -2(8t^2 +5t -3)
divide by -2
0 = (8t^2 +5t -3)
factor
0=(8t-3) (t+1)
using the zero product property
8t-3 = 0 t+1 =0
8t = 3 t= -1
t = 3/8 t= -1
t cannot be negative ( no negative time)
t = 3/8 seconds