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omeli [17]
2 years ago
6

An analog ammeter able to have a 0.5 A maximum reading is to be built using a small coil of resistance R = 20 Ω, and full scale

deflection at 1 mA. What resistance should be added to this coil and how should it be connected to the coil?
Physics
1 answer:
zmey [24]2 years ago
5 0

Answer:

The resistance added to this coil is 0.04 ohm in parallel.

Explanation:

Given that,

Current I'= 0.5 A

Resistance R= 20 ohm

Deflection I= 1 mA

We need to calculate the resistance

Using ohm's law

V = I R

Where,

V = voltage

I =current

R = resistance

V is constant so ,

Therefore,

I R=I'R'

R'=\dfrac{I}{I'}\times R

R'=\dfrac{0.001}{0.5}\times20

R'=0.04\ \Omega

Hence, The resistance added to this coil is 0.04 ohm in parallel.

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A typical human contains 5.00 l of blood, and it takes 1.00 min for all of it to pass through the heart when the person is resti
Mrrafil [7]
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5 0
2 years ago
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A cubical box, 5.00 cm on each side, is immersed in a fluid. The gauge pressure at the top surface of the box is 594 Pa and the
Zolol [24]

Answer:

The density of the fluid is 1100 kg/m³.

Explanation:

Given that,

Height = 5.00 cm

Pressure at top =594 Pa

Pressure at bottom = 1133 Pa

We need to calculate the change in pressure

Using formula of change in pressure

\Delta P=P_{b}-P_{t}

Where, P_{b} = Pressure at bottom

P_{t} = Pressure at top

put the value into the formula

\Delta P=1133-594

\Delta P=539\ Pa

Using formula of pressure for density

\Delta P = \rho g h

\rho =\dfrac{\Delta P}{gh}

Where, \rho = density

P = pressure

h = height

Put the value in to the formula

\rho =\dfrac{539}{5.00\times10^{-2}\times9.8}

\rho =1100\ kg/m^3

Hence, The density of the fluid is 1100 kg/m³.

4 0
2 years ago
A pair of glasses is dropped from the top of a 32.0m stadium. A pen is dropped 2.Os later. How high above the ground is the pen
Svetllana [295]

Answer:

h_p = 30.46\ m

Explanation:

<u>Free Fall Motion</u>

A free-falling object refers to an object that is falling under the sole influence of gravity. If the object is dropped from a certain height h, it moves downwards until it reaches ground level.

The speed vf of the object when a time t has passed is given by:

v_f=g\cdot t

Where g = 9.8 m/s^2

Similarly, the distance y the object has traveled is calculated as follows:

\displaystyle y=\frac{g\cdot t^2}{2}

If we know the height h from which the object was dropped, we can solve the above equation for t:

\displaystyle t=\sqrt{\frac{2\cdot y}{g}}

The stadium is h=32 m high. A pair of glasses is dropped from the top and reaches the ground at a time:

\displaystyle t_1=\sqrt{\frac{2\cdot 32}{9.8}}=2.56\ sec

The pen is dropped 2 seconds after the glasses. When the glasses hit the ground, the pen has been falling for:

t_2=2.56 - 2 = 0.56\ sec

Therefore, it has traveled down a distance:

\displaystyle y=\frac{9.8\cdot 0.56^2}{2} = 1.54\ m

Thus, the height of the pen is:

h_p = 32 - 1.54\Rightarrow h_p=30.46\ m

8 0
2 years ago
A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surfac
GREYUIT [131]

Answer:

The flux through the surface of the cube is 2.314\ Nm^{2}/C

Solution:

As per the question:

Edge of the cube, a = 8.0 cm = 8.0\times 10^{- 2}\ m

Volume Charge density, \rho_{v} = 40 nC/m^{3} = 40\times {- 9}\ C/m^{3}

Now,

To calculate the electric flux:

\phi = \frac{q}{\epsilon_{o}}                                                      (1)

where

\phi = electric flux

\epsilon_{o} = 8.85\times 10^{- 12}\ F/m = permittivity of free space  

Volume Charge density for the given case is given by the formula:

\rho_{v} = \frac{Total\ charge, q}{Volume of cube, V}                  (2)

Volume of cube, V = a^{3}

Thus

V = (8.0\times 10^{- 2})^{3} = 5.12\times 10^{- 4}\ m^{3}

Thus from eqn (2), the total charge is given by:

q = \rho_{v}V = 40\times {- 9}\times 5.12\times 10^{- 4}

q = 2.048\times 10^{-11}\ F = 20.48\ pF

Now, substitute the value of 'q' in eqn (1):

\phi = \frac{2.048\times 10^{-11}}{8.85\times 10^{- 12}} = 2.314\ Nm^{2}/C

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2 years ago
At what condition does a body become weightless at equator
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Hope this is helpful <span>Weightlessness

</span>
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