(a) 138.2 J
Since the applied force is parallel to the displacement of the block, the work done by the force is given by:
where
F = 46.7 N is the magnitude of the force
d = 2.96 m is the displacement of the block
Substituting the numbers into the equation, we find
(b) 45.1 J
In order to calculate the total energy dissipated among the floor and the block as thermal energy, we have to calculate the work done by the frictional force, which is
where
is the coefficient of friction
m = 4.35 kg is the mass of the block
g = 9.8 m/s^2
d = 2.96 m is the displacement
and the negative sign is due to the fact that the frictional force has opposite direction to the displacement.
Substituting, we find
The magnitude of this work is equal to the sum of the thermal energy dissipated between the floor and the block:
And since we know
we find
(c) 58.1 J
According to the work-energy theorem, the increase in kinetic energy of the block must be equal to the work done by the applied force minus the work done by friction (which becomes wasted thermal energy):
Substituting
We find
Answer:
Explanation:
From the question we are told that:
Thickness of plastic membrane 
The temperature of hydrogen 
Diffusion coefficient of hydrogen in the plastic 
The solubility of hydrogen in the plastic membrane 
Generally the equation for molar conc is mathematically given by
3bars
1bar
Generally the equation for molar diffusion flux of Hydrogen N_a is mathematically given by
Answer: 6.4 x 10^14Hz
Explanation:
velocity of blue light (V) = 3.0 x 10^8m/sec
wavelength (λ) = 465 nanometers
(4.65 x 10?m)
Since 1nanometer = 1 x 10^-9 meter
465 nanometer = 4.65 x 10^-7 meters
frequency (F) = ?
Recall that the frequency of a wave is the number of cycles the wave complete in one second, and its unit is Hertz.
So, apply the formula V = F λ
3.0 x 10^8m/sec = F x 4.65 x 10^-7 meters
F = (3.0 x 10^8m/sec / 4.65 x 10^-7 meters)
F = 6.4 x 10^14Hz
Thus, the frequency of the blue light is 6.4 x 10^14Hz
The problem describes the relationship of "bulb a" and "bulb b" to be in connected in series. When the switch is open then no current can flow, on the other hand, when it is closed, current will pass through.
When only "bulb a" is connected to the battery then more current is flowing to "bulb a" causing it to be bright.
Closing the switch would mean that "bulb b" is already included in the circuit and the battery will push small current to flow around the whole circuit. The more bulbs are connected, the harder for the current to flow because the resistance will be very high.
So the light of "bulb a" will be dimmer.
