I think transfers is the answer
Answer:
The rock will reach 9 m from the ground at eaxactly 5.06 s after it was initially thrown upwards.
Explanation:
We will use the equations of motion for this.
u = initial velocity of the rock = 22 m/s
g = acceleration due to gravity = -9.8 m/s²
y = vertical position of the rock at a time t = 9 m
y₀ = initial height of the rock = 25 m
t = time it takes for the rock to reach height of 9 m.
(y-y₀) = ut + 0.5gt²
(9 - 25) = 22t + 0.5(-9.8)t²
- 14 = 22t - 4.9t²
4.9t² - 22t - 14 = 0
solving this quadratic equation,
t = 5.055 s or - 0.565 s
Since time cannot be negative,
t = 5.055 s = 5.06 s
Hope this Helps!!!
Answer:
When we burn wood we are releasing solar energy, in the form of heat, that has been stored in the wood as chemical energy. The process of photosynthesis converted solar energy, water and carbon dioxide into oxygen and the organic molecules that form the wood, half the weight of which is carbon.
Explanation:
Answer:
time required after impact for a puck is 2.18 seconds
Explanation:
given data
mass = 30 g = 0.03 kg
diameter = 100 mm = 0.1 m
thick = 0.1 mm = 1 ×
m
dynamic viscosity = 1.75 ×
Ns/m²
air temperature = 15°C
to find out
time required after impact for a puck to lose 10%
solution
we know velocity varies here 0 to v
we consider here initial velocity = v
so final velocity = 0.9v
so change in velocity is du = v
and clearance dy = h
and shear stress acting on surface is here express as
= µ 
so
= µ
............1
put here value
= 1.75×
× 
= 0.175 v
and
area between air and puck is given by
Area =
area =
area = 7.85 ×
m²
so
force on puck is express as
Force = × area
force = 0.175 v × 7.85 × 
force = 1.374 ×
v
and now apply newton second law
force = mass × acceleration
- force = 
- 1.374 ×
v = 
t = 
time = 2.18
so time required after impact for a puck is 2.18 seconds