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2 years ago

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12*8A hollow steel ball weighing 4 pounds is suspended from a spring. This stretches the spring 13 feet. The ball is started in

motion from the equilibrium position with a downward velocity of 2 feet per second. The air resistance (in pounds) of the moving ball numerically equals 4 times its velocity (in feet per second) . Suppose that after t seconds the ball is y feet below its rest position. Find y in terms of t. (Note that the positive direction is down.) Take as the gravitational acceleration 32 feet per second per second

1 answer:

max2010maxim [7]2 years ago

6 0

Answer:

See attached pictures.

Explanation:

See attachments for explanation.

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Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

balandron [24]

Answer with Explanation:

We are given that

Radius of solid core wire=r=2.28 mm= $2.28\times 10^{-3} m$

$1mm=10^{-3} m$

Radius of each strand of thin wire=r'=0.456 mm= $0.456\times 10^{-3} m$

Current density of each wire= $J=3750 A/m^2$

a.Area = $\pi r^2$

Where $\pi=3.14$

Using the formula

Cross section area of copper wire has solid core = $3.14\times (2.28\times 10^{-3})^2=16.3\times 10^{-6} m^2$

Current density = $J=\frac{I}{A}$

Using the formula

$3750=\frac{I}{16.3\times 10^{-6}}$

$I=3750\times 16.3\times 10^{-6}=0.061 A$

Total number of strands=19

Area of strand wire= $A'=19\times 3.14\times (0.456\times 10^{-3})^2=12.4\times 10^{-6} m^2$

$J'=\frac{I'}{A'}$

$3750=\frac{I'}{19\times 3.14(0.456\times 10^{-3})^2}$

$I'=3750\times 19\times 3.14(0.456\times 10^{-3})^2$

$I'=0.047 A$

b.Resistivity of copper wire= $\rho=1.69\times 10^{-8}\Omega-m$

Length of each wire =6.25 m

Resistance, R= $\frac{\rho l}{A}$

Using the formula

Resistance of solid core wire= $R=\frac{1.69\times 10^{-8}\times 6.25}{16.3\times 10^{-6}}=6.5\times 10^{-3}\Omega$

Resistance of strand wire= $R'=\frac{1.69\times 10^{-8}\times 6.25}{12.4\times 10^{-6}}=8.5\times 10^{-3}\Omega$

7 0

2 years ago

An electron is moving in the vicinity of a long, straight wire that lies along the z-axis. The wire has a constant current of 8.

viktelen [127]

Answer:

The force that the wire exerts on the electron is $-4.128\times10^{-20}i-6.88\times10^{-20}j+0k$

Explanation:

Given that,

Current = 8.60 A

Velocity of electron $v= (5.00\times10^{4})i-(3.00\times10^{4})j\ m/s$

Position of electron = (0,0.200,0)

We need to calculate the magnetic field

Using formula of magnetic field

$B=\dfrac{\mu I}{2\pi d}(-k)$

Put the value into the formula

$B=\dfrac{4\pi\times10^{-7}\times8.60}{2\pi\times0.200}$

$B=0.0000086\ T$

$B=-8.6\times10^{-6}k\ T$

We need to calculate the force that the wire exerts on the electron

Using formula of force

$F=q(\vec{v}\times\vec{B}$

$F=1.6\times10^{-6}((5.00\times10^{4})i-(3.00\times10^{4})j\times(-8.6\times10^{-6}) )$

$F=(1.6\times10^{-19}\times3.00\times10^{4}\times(-8.6\times10^{-6}))i+(1.6\times10^{-19}\times5.00\times10^{4}\times(-8.6\times10^{-6}))j+0k$

$F=-4.128\times10^{-20}i-6.88\times10^{-20}j+0k$

Hence, The force that the wire exerts on the electron is $-4.128\times10^{-20}i-6.88\times10^{-20}j+0k$

5 0

2 years ago

Susie walks 3 blocks north to the local CVS store, then 4 blocks east to her grandmother’s house. She then walks 2 blocks west a

Slav-nsk [51]

Answer:

Suzie is 3 blocks north of where she started

Explanation:

Displacement is the minimum distance between the initial and final point of motion.

Here, Suzie first walks 3 blocks north. From there she walks 4 blocks east. Then 2 blocks to the east then 2 blocks north and then 2 blocks east. She covered 4 blocks east toward west. This is the same distance she covered traveling east. But she is 2 blocks north. From there she traveled a block south to the pizzeria and another block to her friends house. She covered the two block she had traveled north.

Hence, Suzie is 3 blocks north of where she started.

7 0

2 years ago

Consider a star that is a sphere with a radius of 6.32 108 m and an average surface temperature of 5350 K. Determine the amount

Mariulka [41]

Answer:

The value is $\Delta s = 8.537 *10^{25 } \ J/K$

Explanation:

From the we are told that

The radius of the sphere is $r = 6.32 *10^{8} \ m$

The temperature is $T_x = 5350 \ K$

The average temperature of the rest of the universe is $T_r = 2.73 \ K$

Generally the change in entropy of the entire universe per second is mathematically represented as

$\Delta s = s_r - s_x$

Here $s_r$ is the entropy of the rest of the universe which is mathematically represented as

$s_r = \frac{Q}{T_r}$

Here Q is the quantity of heat radiated by the star which is mathematically represented as

$Q = 4 \pi * r^2 * \sigma * T^4_x$

Here $\sigma$ is the Stefan-Boltzmann constant with value

$\sigma = 5.67 * 10^{-8 }W\cdot m^{-2} \cdot K^{-4}.$

=> $Q = 4 \pi * (6.32*10^{8})^2 * 5.67 * 10^{-8 } * 5350 ^4$

=> $Q = 2.332 *10^{26} \ J$

So

$s_r = \frac{2.332 *10^{26}}{2.73}$

=> $s_r = 8.5415 *10^{25}\ J/K$

Here $s_x$ is the entropy of the rest of the universe which is mathematically represented as

$s_x = \frac{Q}{T_x}$

=> $s_x = \frac{2.332 *10^{26} }{5350}$

=> $s_x = 4.359 *10^{22} \ J/K$

So

$\Delta s = 8.5415 *10^{25} - 4.359 *10^{22}$

=> $\Delta s = 8.537 *10^{25 } \ J/K$

7 0

2 years ago

It requires 49 J of work to stretch an ideal very light spring from a length of 1.4 m to a length of 2.9 m. What is the value of

nadya68 [22]

Answer:

44 N/m

Explanation:

The extension, e, of the spring = 2.9 m - 1.4 m = 1.5 m

The work needed to stretch a spring by <em>e</em> is given by

$W = \frac{1}{2} ke^2$

where <em>k</em> is spring constant.

$k = \dfrac{2W}{e^2}$

Using the appropriate values,

$k = \dfrac{2\times 49\text{ J}}{1.5^2\text{ m}^2} = 43.55\ldots\text{ N/m} \approx 44\text{ N/m}$

3 0

2 years ago

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