Force applied by the machine to over come resistance
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.
The magnification of the ornament is 0.25
To calculate the magnification of the ornament, first, we need to find the image distance.
Formula:
- 1/f = u⁻¹+v⁻¹.................... Equation 1
Where:
- f = Focal length of the ornament
- u = image distance
- v = object distance.
make u the subject of the equation
- u = fv/(f+v)................ Equation 2
From the question,
Given:
Substitute these values into equation 2
- u = (12×4)/(12+4)
- u = 48/16
- u = 3 cm.
Finally, to get the magnification of the ornament, we use the formula below.
- M = u/v.................. Equation 3
Where
- M = magnification of the ornament.
Substitute these values above into equation 3
Hence, The magnification of the ornament is 0.25
Answer:
Here
Length (l) = 6 m
height (h) = 3 m
Load(L) = 500 N
Effort (E) = ?
we know the principal that
E * l = L * h
6 E = 500 * 3
6E = 1500
E = 250
therefore 250 N work is done on the barrel.
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