Answer:
Increase in surface area leads to more chances for connections between neurons hence more possible learning
Explanation:
The brain appears wrinkled due to the presence of gyri and sulci. Sulci are the grooves in the brain while gyri are the folds. The two major functions of gyri and sulci are to lead to formation of brain divisions and increase the surface area of cerebral cortex. The formation of brain divisions is done by creating boundaries between the brain lobes thereby leading to division of the brain into two hemispheres. Moreover, by increasing the surface area of the brain, the cortex processes more information since more neurons are formed.
Answer:
Galileo Galilei
Explanation:
although Galileo was not the scientist who invented the telescope, he was the first to use it to observe celestial objects. he used the telescope in 1609. his discovery included more accurate information about the moon, the sun and some of the planets.
Answer:
a) A = 0.603 m
, b) a = 165.8 m / s²
, c) F = 331.7 N
Explanation:
For this exercise we use the law of conservation of energy
Starting point before touching the spring
Em₀ = K = ½ m v²
End Point with fully compressed spring
= 
= ½ k x²
Emo =
½ m v² = ½ k x²
x = √(m / k) v
x = √ (2.00 / 550) 10.0
x = 0.603 m
This is the maximum compression corresponding to the range of motion
A = 0.603 m
b) Let's write Newton's second law at the point of maximum compression
F = m a
k x = ma
a = k / m x
a = 550 / 2.00 0.603
a = 165.8 m / s²
With direction to the right (positive)
c) The value of the elastic force, let's calculate
F = k x
F = 550 0.603
F = 331.65 N
A. Since it does not say the ball is moving your answer is A.
Answer:
Tension= 21,900N
Components of Normal force
Fnx= 17900N
Fny= 22700N
FN= 28900N
Explanation:
Tension in the cable is calculated by:
Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium
FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)
Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)
Ftorque= 2/3FBcostheta+ 4/3FWcostheta
Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°
Ftorque= 21900N
b) components of Normal force
Efx=FNx-FTcos(90-theta)=0 static equilibrium
Fnx=21900cos(90-55)=17900N
Fy=FNy+ FTsin(90-theta)-FB-FW=0
FNy= -FTsin(90-55)+FB+FW
FNy= -21900sin(35)+(1350+2250)×9.81=22700N
The Normal force
FN=sqrt(17900^2+22700^2)
FN= 28.900N
