Is the Frog Toy a piece of literature? If so, the type of wave is the "Wave of Life." Spring brings new life and a new start from the harsh winter.
Answer:
a)Distance traveled during the first second = 4.905 m.
b)Final velocity at which the object hits the ground = 38.36 m/s
c)Distance traveled during the last second of motion before hitting the ground = 33.45 m
Explanation:
a) We have equation of motion
S = ut + 0.5at²
Here u = 0, and a = g
S = 0.5gt²
Distance traveled during the first second ( t =1 )
S = 0.5 x 9.81 x 1² = 4.905 m
Distance traveled during the first second = 4.905 m.
b) We have equation of motion
v² = u² + 2as
Here u = 0, s= 75 m and a = g
v² = 0² + 2 x g x 75 = 150 x 9.81
v = 38.36 m/s
Final velocity at which the object hits the ground = 38.36 m/s
c) We have S = 0.5gt²
75 = 0.5 x 9.81 x t²
t = 3.91 s
We need to find distance traveled last second
That is
S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m
Distance traveled during the last second of motion before hitting the ground = 33.45 m
Answer:
<h2>3.36J</h2>
Explanation:
Step one:
given data
mass m= 1.3kg
distance moved s= 2.8m
opposing frictional force= 0.34N
assume g= 9.81m/s^2
we know that work done= force *distance moved
1. work done to push the book= 1.55*2.8=4.34J
2. Work against friction = force of friction x distance
= 0.34*2.8=0.952J
Step two:
the work done on the book is the net work, which is
Network done= work done to push the book- Work against friction
Network done= 4.32-0.952=3.36J
<u>Therefore the work of the 1.55N 3.36J</u>
