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irga5000 [103]
3 years ago
12

what factors limit the accuracy of a potentiometer and what was the objective of reversing the terminals of the cell​

Physics
1 answer:
cricket20 [7]3 years ago
7 0

Explanation:

<em>"The accuracy of a potentiometer can be increased by decreasing the potential gradient across the potentiometer wire, and this can be achieved by increasing the length"</em>

<em />

<u>The factors that are affecting/limiting the accuracy of the potentiometer are: </u>

  1. The specific resistance of the material of the potentiometer wire.
  2. The potential gradient
  3. The current passing through the potentiometer wire.
  4. Area of a cross-section of the wire
  5. Internal temperature.

     

<u>The objective of reversing the terminals of the cell​</u>

If the jockey of the potentiometer is pressed for a long time, joule heating sets in, so that reversing the terminals of the potentiometer will prevent the resistance due to joule heat from being added to the measured resistance, ultimately preventing unwanted resistance

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A 8.34 × 103-kg lunar landing craft is about to touch down on the surface of the moon, where the acceleration due to gravity is
Mnenie [13.5K]

Answer:

21870.3156 N

Explanation:

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 1.6 m/s²

Equation of motion

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-18.2^2}{2\times 162}\\\Rightarrow a=-1.02234\ m/s^2

The acceleration of the craft should be 1.02234 m/s²

F=ma\\\Rightarrow F=8.34\times 10^3\times 1.02234\\\Rightarrow F=8526.3156\ N

Weight of the craft

W=mg\\\Rightarrow W=8.34\times 10^3\times 1.6\\\Rightarrow W=13344\ N

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F_t=F+W\\\Rightarrow F_t=8526.3156+13344\\\Rightarrow F_t=21870.3156\ N

The thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface is 21870.3156 N

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3 years ago
How much potintial energy is in a closed system where the height of the object is 2 m, and the mass is 10 kg?
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Answer:

D. 196 J

Explanation:

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Which increases the work output of a machine?
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A reducing friction
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The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of
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Answer: 230.50 m

Explanation:

We have the following information:

h_{Hg-TOP}=675mmHg=0.675m the barometric reading at the top of the building

h_{Hg-BOT}=695mmHg=0.695m the barometric reading at the bottom of the building

\rho _{air}=1.18 kg/m^{3} density of air

\rho _{Hg}=13600 kg/m^{3} density of mercury

g=9.8/m^{2} gravity

And we need to find the height of the building.

In order to approach this problem, we will firstly use the following equations to find the pressure at the top of the building P_{TOP} and the perssure at the bottom P_{BOT}:

P_{TOP}=\rho _{Hg} g h_{Hg-TOP} (1)

P_{BOT}=\rho _{Hg} g h_{Hg-BOT} (2)

From (1): P_{TOP}=(13600 kg/m^{3})(9.8/m^{2})(0.675m)=89964 Pa (3)

From (2): P_{BOT}=(13600 kg/m^{3})(9.8/m^{2})(0.695m)=92629.6 Pa (4)

Having the pressures at the top and the bottom of the building, we can calculate the variation in pressure \Delta P:

\Delta P=P_{BOT} - P_{TOP} (5)

\Delta P=92629.6 Pa - 89964 Pa=2665.6 Pa (6)

On the other hand, we have a column of air with a cross-section area A and the same height of the building, lets name it h_{air}.

As pressure is defined as the force F exerted on a specific area A, we can write:

\Delta P=\frac{F}{A} (7)

If we isolate F we have:

F= A \Delta P (8)

Also, the force gravity exerts on this column of air (its weight) is:

F=m_{air} g (9)

Knowing the density of air is: \rho_{air}=\frac{m_{air}}{V_{air}} (10)

where the volume of air can be written as: V_{air}=(A)(h_{air}) (11)

Substituting (1) in (10):

\rho_{air}=\frac{m_{air}}{(A)(h_{air}} (12)

Isolating m_{air}:

m_{air}=(\rho_{air}) (A) (h_{air}) (13)

Substituting (13) in (9):

F=(\rho_{air}) (A) (h_{air}) (g) (14)

Matching (8) and (14)

A \Delta P=(\rho_{air}) (A) (h_{air}) (g) (15)

Isolating h_{air}:

h_{air}=\frac{\Delta P}{g \rho_{air}} (16)

Substituting the known and calculated values:

h_{air}=\frac{2665.6 Pa}{(9.8m/s^{2}) (1.18 kg/m^{3})} (17)

Finally:

h_{air}=230.50 m This is the height of the building

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Answer:

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