Answer:
See attached document
Explanation:
Entire process for deriving the asked expression dV across the bridge as function of dP is illustrated in the attachment below.
The document gives a step-by step process for arriving at the expression. However, manipulation of algebraic equations is skipped for the conciseness of the document.
It also gives the expression for the case when all resistors have different nominal values.
Answer:
Push with force of 1N
Explanation:
I have explained in the paper.
Goodluck
(a) The velocity (in m/s) of the rock after 1 second is 11.28 m/s.
(b) The velocity of the rock after 2 seconds is 7.56 m/s.
(c) The time for the block to hit the surface is 4.03.
(d) The velocity of the block at the maximum height is 0.
<h3>
Velocity of the rock</h3>
The velocity of the rock is determined as shown below;
Height of the rock after 1 second; H(t) = 15(1) - 1.86(1)² = 13.14 m
v² = u² - 2gh
where;
- g is acceleration due to gravity in mars = 3.72 m/s²
v² = (15)² - 2(3.72)(13.14)
v² = 127.23
v = √127.23
v = 11.28 m/s
<h3>Velocity of the rock when t = 2 second</h3>
v = dh/dt
v = 15 - 3.72t
v(2) = 15 - 3.72(2)
v(2) = 7.56 m/s
<h3>Time for the rock to reach maximum height</h3>
dh/dt = 0
15 - 3.72t = 0
t = 4.03 s
<h3>Velocity of the rock when it hits the surface</h3>
v = u - gt
v = 15 - 3.72(4.03)
v = 0
Learn more about velocity at maximum height here: brainly.com/question/14638187
Since the electric field between the plates is constant, If the two plates are brought closer together, the potential difference between the two plates decreases
The relation between potential difference and the electric field is given by ΔV = E.d
Since the electric field is maintained constant, the potential difference is directly inversely proportional to the distance between the plates.
The potential difference between the plates will therefore likewise decrease if the distance between the plates is reduced, we will state in this case.
The energy required to move a unit charge, or one coulomb, from one point to the other in a circuit is measured as the potential difference between the two points. Potential difference is measured in volts or joules per coulomb.
Refer to more about the potential difference here
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Answer:
Recessed incandescent luminaires not marked type ic and those marked for installing directly in insulated ceilings must not have insulation over the top of the luminaire.
Explanation:
Depending on how they interact with insulation, lighting fixtures are rated at various levels. Non-IC rated lighting fixtures can accommodate higher wattage bulbs, but they also pose the greatest fire risk when used with the incorrect insulation.
In locations with insulation, light fixtures that are not IC rated may be installed. But there is a condition. The distance between the fixture and any insulation should be 3 inches. But the 3 inch gap in the insulation would negate the goal of insulation by producing a lot of uninsulated space, so this defies logic. Building a box-style cover to cover the fixture on the attic side is one option to fix this. Drywall or foil-faced foam insulation can be used to create this box. After the cover is put in place, insulation can be added for maximum effectiveness.
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