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ss7ja [257]
4 years ago
8

What should a voltmeter that is attached to three batteries, each measuring 1.5 volts read:

Physics
1 answer:
Pie4 years ago
4 0

Answer:

The answer depends on the arrangement of the batteries either parallel or series.

If the three batteries are connected in parallel, the voltmeter will read the same voltage as the individual battery (I.e, the combine voltage will remain 1.5 Bolts)

While

If the batteries are connected in series, and are correctly connected together from positive to negative, the combine voltages will increase by adding individual battery voltage together. So for three 1.5 Volts batteries, the total voltages will be 4.5 Volts.

Explanation:

Parallel connected

Vp = 1.5 Volts

Series connection

Vs = 1.5 + 1.5 + 1.5 = 4.5Volts

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A technician is troubleshooting a problem. The technician tests the theory and determines the theory is confirmed. Which of the
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_____________ determines if an object will float or sink in a fluid.
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Density.

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A river flows due east at 1.60 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v
Citrus2011 [14]

Answer:

part (a) v\ =\ 10.42\ at\ 81.17^o towards north east direction.

part (b) s = 46.60 m

Explanation:

Given,

  • velocity of the river due to east = v_r\ =\ 1.60\ m/s.
  • velocity of the boat due to the north = v_b\ =\ 10.3\ m/s.

part (a)

River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,

Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are 90^o

Let 'v' be the velocity of the boat relative to the shore.

\therefore v\ =\ \sqrt{v_r^2\ +\ v_b^2}\\\Rightarrow v\ =\ \sqrt{1.60^2\ +\ 10.3^2}\\\Rightarrow v\ =\ 10.42\ m/s.

Let \theta be the angle of the velocity of the boat relative to the shore with the horizontal axis.

Direction of the velocity of the boat relative to the shore.\therefore Tan\theta\ =\ \dfrac{v_b}{v_r}\\\Rightarrow Tan\theta\ =\ \dfrac{10.3}{1.60}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{10.3}{1.60}\ \right )\\\Rightarrow \theta\ =\ 81.17^o

part (b)

  • Width of the shore = w = 300m

total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken

Let 't' be the total time taken by the boat to cross the width of the river.\therefore w\ =\ v_bt\\\Rightarrow t\ =\ \dfrac{w}{v_b}\\\Rightarrow t\ =\ \dfrac{300}{10.3}\\\Rightarrow t\ =\ 29.12 s

Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river\therefore s\ =\ u_rt\\\Rightarrow s\ =\ 1.60\times 29.12\\\Rightarrow s\ =\ 46.60\ m

7 0
4 years ago
A mass with mass 4 is attached to a spring with spring constant 24 and a dashpot giving a damping 20. The mass is set in motion
aleksandrvk [35]

Answer: x(t) = 14e^{-2t} - 10e^{-3t}

Explanation: In a mass-spring-damper system, the differential equation that rules the motion of the mass is: mx" + cx' + kx = 0

Using m = 4, k = 24 and c = 20, we have

4x" + 20x' + 24x = 0

Simplifying, we have

x" + 5x'+ 6x = 0

The characteristic equation of this differential is

r^{2} + 5r + 6 = 0

The solutions for the quadratic equation are: r_{1} = -2 and r_{2} = -3

Hence:

x(t) = C_{1}e^{-2t} + C_{2}e^{-3t}

x'(t) = -2C_{1}E^{-2t} - 3C_{2}e^{-3t}

To determine the constants, we have the initial conditions x(0) = 4 and

x'(0) = 2, then:

x(0) = C_{1} + C_{2} = 4\\          C_{1} = 4 - C_{2}

x'(0) = -2C_{1} -3C_{2} = 2\\-2(4-C_{2}) -3C_{2} = 2\\C_{2} = -10\\C_{1} = 4 - C_{2}\\C_{1} = 14

Substituing the constants:

x(t) = 14e^{-2t} - 10e^{-3t}

The position function for this system is: x(t) = 14e^{-2t} - 10e^{-3t}

7 0
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