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Kruka [31]
3 years ago
10

If the mass of both weights is 225 gm, the first mass is located 20∘ north of east, the second mass is located 20∘ south of east

, and the transducer sensitivity is 0.5 volts/Newton, how large a voltage do you expect to measure? Assume the transducer has been properly zeroed so that V = 0 when F3=0. Please express your answers with 1 decimal place.
Physics
1 answer:
Montano1993 [528]3 years ago
7 0

Answer:

The voltage is 2.114 V.

Explanation:

Given that,

Mass of both weights = 225 gm

Transducer sensitivity = 0.5 V/N

The first mass is located 20∘ north of east, the second mass is located 20∘ south of east,

We need to calculate the net equivalent force

Using formula of force

F_{3}=m_{1}g\cos\theta+m_{2}g\cos\theta

F_{3}=2mg\cos\theta

Put the value into the formula

F_{3}=2\times0.225\times10\cos20^{\circ}

F_{3}=4.228\ N

We need to calculate the voltage

Using formula of voltage

Voltage =sensitivity\times F_{3}

Put the value into the formula

Voltage=0.5\times4.228

Voltage =2.114\ V

Hence, The voltage is 2.114 V.

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Answer:

The value of  charge q₃ is 40.46 μC.

Explanation:

Given that.

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Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

r=\sqrt{x_{2}^2+y_{2}^2}

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r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}

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Using formula of net force

F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})

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14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})

(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}

\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}

q_{3}=0.0210909\times(0.0438)^2

q_{3}=40.46\times10^{-6}\ C

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Nitella [24]

Answer:

Total number of lamps will be 4            

Explanation:

We have given power of the lamp W = 400 watt

Potential difference across the lamp V=110 volt

We know that power is equal to P=VI

So 400=110\times I

I=3.636A

Total current is given 15 A

As it is given that lamps are connected in parallel so total current is the sum of current through each lamp

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3 years ago
a chamber with a fixed volume of 1.0 meters cubed contains a monatomic gas at 3.00 *10^K. The chamber is heated to a temperature
igomit [66]

Answer:

Explanation:

Given

Volume of fixed chamber V=1 m^3

Initial Temperature T_1=300 K

Final Temperature T_2=400 K

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Change in internal energy of the system is equal to heat added minus work done by the system

\Delta U=Q-W

as the volume is fixed therefore work

W=\int PdV=0

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n\times 12.471\times (400-300)=10

n=0.008018 mol

and 1 mole contains 6.022\times 10^{23} molecules

thus  No of molecules=0.008018\times 6.022\times 10^{23}

No of molecules=4.82\times 10^{21} molecules

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3 years ago
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