A test car and its driver, with a combined mass of 600 kg, are moving along a straight,horizontal track when a malfunction cause
1 answer:
You might be interested in
Answer:
a) 8.99*10³ V b) 4.5*10⁻² J c) 0 d) 0
Explanation:
a)
- The electrostatic potential V, is the work done per unit charge, by the electrostatic force, producing a displacement d from infinity (assumed to be the reference zero level).
- For a point charge, it can be expressed as follows:
- As the electrostatic force is linear with the charge (it is raised to first power), we can apply superposition principle.
- This means that the total potential at a given point, is just the sum of the individual potentials due to the different charges, as if the others were not there.
- In our case, due to symmetry, the potential, at any corner of the triangle, is just the double of the potential due to the charge located at any other corner, as follows:
- The potential at point C is 8.99*10³ V
b)
- The work required to bring a positive charge of 5μC from infinity to the point C, is just the product of the potential at this point times the charge, as follows:
- The work needed is 0.045 J.
c)
- If we replace one of the charges creating the potential at the point C, by one of the same magnitude, but opposite sign, we will have the following equation:
- This means that the potential due to both charges is 0, at point C.
d)
- If the potential at point C is 0, assuming that at infinity V=0 also, we conclude that there is no work required to bring the charge of 5μC from infinity to the point C, as no potential difference exists between both points.
Explanation :
In transverse waves the particles are oscillating perpendicular to the direction of propagation of waves.
The uppermost part of the wave is crests and the lowermost part is troughs.
Wavelength of a transverse wave is defined as the distance between two consecutive crests or troughs.
Amplitude is the maximum distance or displacement covered by a wave.
So, crest, amplitude, trough and wavelength identifies the parts of a transverse wave.
Answer:
amount of energy = 4730.4 kWh/yr
amount of money = 520.34 per year
payback period = 0.188 year
Explanation:
given data
light fixtures = 6
lamp = 4
power = 60 W
average use = 3 h a day
price of electricity = $0.11/kWh
to find out
the amount of energy and money that will be saved and simple payback period if the purchase price of the sensor is $32 and it takes 1 h to install it at a cost of $66
solution
we find energy saving by difference in time the light were
ΔE = no of fixture × number of lamp × power of each lamp × Δt
ΔE is amount of energy save and Δt is time difference
so
ΔE = 6 × 4 × 365 ( 12 - 9 )
ΔE = 4730.4 kWh/yr
and
money saving find out by energy saving and unit cost that i s
ΔM = ΔE × Munit
ΔM = 4730.4 × 0.11
ΔM = 520.34 per year
and
payback period is calculate as
payback period =
payback period =
payback period = 0.188 year
1) Current in the wire: 0.0875 A
The current in the wire is given by:
where
Q is the charge passing a given point in the conductor
t is the time elapsed
In this problem, we have
Q = 420 C is the total charge passing through a given point in a time of
t = 80 min = 4800 s
So, the current is
2) Drift velocity of the electrons:
The drift velocity of the electrons in the wire is given by:
where
I = 0.0875 A is the current
is the number of free electrons per cubic meter
A is the cross-sectional area
is the charge of one electron
The radius of the wire is
So the cross-sectional area is
So, the drift velocity is