<u>Answer:</u>
<em>1. A NaCl solution with a concentration of 50g/100mL of water at 40°C:</em> The NaCl solution with a given concentration is saturated at this temperature .As the temperature increases the solution will more dissolves.
<em>2. A sugar solution with a concentration of 200g/100mL of water at 40°C: </em>The sugar solution with a given concentration is saturated at this temperature. As the temperature increases the solution will more dissolves.
<em>3. A sugar solution with a concentration of 240g/100mL of water at 40°C:</em> The sugar solution with a given concentration is saturated at given temperature.
The vesicles release neurotransmitters. These cross the synapse and are accepted by the receptors in the dendrites of the next neuron.
Explanation:
An axon, or nerve fiber, is a long slender projection of a nerve cell, or neuron, that conducts electrical impulses away from the neuron's cell body. Axons are in effect the primary transmission lines of the nervous system, and as bundles they help make up nerves.
When an action potential reaches the axon terminal, it depolarizes the membrane and opens voltage-gated Na+ channels. Na+ ions enter the cell, further depolarizing the presynaptic membrane.
<h2>
Answer:</h2>
(a) 3.96 x 10⁵C
(b) 4.752 x 10⁶ J
<h2>
Explanation:</h2>
(a) The given charge (Q) is 110 A·h (ampere hour)
Converting this to A·s (ampere second) gives the number of coulombs the charge represents. This is done as follows;
=> Q = 110A·h
=> Q = 110 x 1A x 1h [1 hour = 3600 seconds]
=> Q = 110 x A x 3600s
=> Q = 396000A·s
=> Q = 3.96 x 10⁵A·s = 3.96 x 10⁵C
Therefore, the number of coulombs of charge is 3.96 x 10⁵C
(b) The energy (E) involved in the process is given by;
E = Q x V -----------------(i)
Where;
Q = magnitude of the charge = 3.96 x 10⁵C
V = electric potential = 12V
Substitute these values into equation (i) as follows;
E = 3.96 x 10⁵ x 12
E = 47.52 x 10⁵ J
E = 4.752 x 10⁶ J
Therefore, the amount of energy involved is 4.752 x 10⁶ J
(a) 5.66 m/s
The flow rate of the water in the pipe is given by
where
Q is the flow rate
A is the cross-sectional area of the pipe
v is the speed of the water
Here we have
the radius of the pipe is
r = 0.260 m
So the cross-sectional area is
So we can re-arrange the equation to find the speed of the water:
(b) 0.326 m
The flow rate along the pipe is conserved, so we can write:
where we have
and where 
is the cross-sectional area of the pipe at the second point.
Solving for A2,
And finally we can find the radius of the pipe at that point:
Answer:
Ff = 19.6 N
Explanation:
So since its saying whats the minimum F to move the block, we will use static friction (0.5).
We will use the equation for force of friction, which is Ff = uFn
Ff = (0.5)(4)(9.8)
Ff = 19.6 N
this is the minumum force needed to move the block, as that is the frictional force. You would need to apply a minimum force of 19.6 N to move the block
