Transpiration is the what the process is called.
It controls the release of water through the stomata.
Answer:
The Correct Answer is Heat of Combustion
Explanation:
Answer:
A = -14.87 i ^ + 8.42 j ^ + 0 k ^
B = -25.41 i ^ -12.0 j ^ + 0 k ^
Explanation:
For this exercise let's use trigonometry by decomposing to vectors
vector A
module 17.1 with an angle of 150.5 counterclockwise.
Sin 150.5 =
/ A
cos 150.5 = Ax / A
A_{y} = A sin 150.5 = 17.1 sin 150.5
Aₓ = A cos 1505 = 172 cos 150.5
A_{y} = 8,420
Aₓ = -14.870
the vector is
A = -14.87 i ^ + 8.42 j ^ + 0 k ^
Vector B
= 28.1 sin 205.3
Bₓ = 28.1 cos 205.3
B_{y} = -12.009
Bₓ = -25.405
the vector is
B = -25.41 i ^ -12.0 j ^ + 0 k ^
Answer:
459.6J
Explanation:
Given parameters:
Angle of pull = 40°
Force applied = 30N
Distance moved = 20m
Unknown:
Work done by Kraig = ?
Solution:
To solve this problem;
Work done = F x dcosФ
d is the distance
F is the force
Ф is the angle given
Work done = 30 x 20cos40° = 459.6J
Answer:
Explanation:
The question relates to motion on a circular path .
Let the radius of the circular path be R .
The centripetal force for circular motion is provided by frictional force
frictional force is equal to μmg , where μ is coefficient of friction and mg is weight
Equating cenrtipetal force and frictionl force in the case of car A
mv² / R = μmg
R = v² /μg
= 26.8 x 26.8 / .335 x 9.8
= 218.77 m
In case of moton of car B
mv² / R = μmg
v² = μRg
= .683 x 218.77x 9.8
= 1464.35
v = 38.26 m /s .