Answer:
Corpus callosum
Explanation:
The corpus callosum is the tissue in the brain that helps in the connection between the right and left cerebral hemisphere.
The corpus callosum consists of about 200 million axons of thick nerve cells which helps in combining sensory and cognitive signals between the two hemispheres.
Almost everyone is born with this tissue because an absence of deficiency of it results in serious neurological problems such as difficulty in learning and other cognitive related diseases.
Answer : The partial pressure of
is, 67.009 atm
Solution : Given,
Partial pressure of
at equilibrium = 30.6 atm
Partial pressure of
at equilibrium = 13.9 atm
Equilibrium constant = 
The given balanced equilibrium reaction is,

The expression of
will be,

Now put all the values of partial pressure, we get


Therefore, the partial pressure of
is, 67.009 atm
Answer:
t = 0.319 s
Explanation:
With the sudden movement of the athlete a pulse is formed that takes time to move along the rope, the speed of the rope is given by
v = √T/λ
Linear density is
λ = m / L
λ = 4/20
λ = 0.2 kg / m
The tension in the rope is equal to the athlete's weight, suppose it has a mass of m = 80 kg
T = W = mg
T = 80 9.8
T = 784 N
The pulse rate is
v = √(784 / 0.2)
v = 62.6 m / s
The time it takes to reach the hook can be searched with kinematics
v = x / t
t = x / v
t = 20 / 62.6
t = 0.319 s
Answer:
The value is 
Explanation:
From the question we are told that
The focal length of the objective is 
The focal length of the eyepiece is 
The tube length is 
Generally the magnitude of the overall magnification is mathematically represented as

Where
is the objective magnification which is mathematically represented as

=> 
=> 
is the eyepiece magnification which is mathematically evaluated as



So


Answer:
375 ms
Explanation:
the frequency of metronome , f = 160 beats per minute
f = 160 /60 beats per sec
f = 2.67 beats /s
the period of a single beat , T = 1/f
T = 1/2.67 s
T = 0.375 s = 375 ms
the period of a single beat is 375 ms