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Natasha2012 [34]

2 years ago

5

Two solenoids of equal length are each made of 2000 turns of copper wire per meter. Solenoid I has a 5.00 cm radius; solenoid II

a 10.0 cm radius. When equal currents are present in the two solenoids, the ratio of the magnitude of the magnetic field BIalong the axis of solenoid I to the magnitude of the magnetic field BIIalong the axis of solenoid II, BI/BII, is

1 answer:

Nookie1986 [14]2 years ago

5 0

Answer:

BI/BII = 1

Explanation:

The magnetic field due to a solenoid is given by the following formula:

$B = \mu nI\\$

where,

B = Magnetic Field due to solenoid

μ = permeability of free space

n = No. of turns per unit length

I = current passing through the solenoid

Now for the first solenoid:

$B_1 = \mu n_1I_1 \\$

For the second solenoid:

$B_2 = \mu n_2I_2\\$

Dividing both equations:

$\frac{B_1}{B_2} = \frac{\mu n_1I_1}{\mu n_2I_2}\\$

here, no. of turns and the current passing through each solenoid is same:

n₁ = n₂ and I₁ = I₂

Therefore,

$\frac{B_1}{B_2} = \frac{\mu nI}{\mu nI}\\$

<u>BI/BII = 1</u>

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Fiesta28 [93]

Answer:

$X_2=25.27m$

Explanation:

Here we will call:

1. $E_1$ : The energy when the first spring is compress

2. $E_2$ : The energy after the mass is liberated by the spring

3. $E_3$ : The energy before the second string catch the mass

4. $E_4$ : The energy when the second sping compressed

so, the law of the conservations of energy says that:

1. $E_1 = E_2$

2. $E_2 -E_3= W_f$

3. $E_3 = E_4$

where $W_f$ is the work of the friction.

1. equation 1 is equal to:

$\frac{1}{2}Kx^2 = \frac{1}{2}MV_2^2$

where K is the constant of the spring, x is the distance compressed, M is the mass and $V_2$ the velocity, so:

$\frac{1}{2}(20)(8)^2 = \frac{1}{2}(0.3)V_2^2$

Solving for velocity, we get:

$V_2$ = 65.319 m/s

2. Now, equation 2 is equal to:

$\frac{1}{2}MV_2^2-\frac{1}{2}MV_3^2 = U_kNd$

where M is the mass, $V_2$ the velocity in the situation 2, $V_3$ is the velocity in the situation 3, $U_k$ is the coefficient of the friction, N the normal force and d the distance, so:

$\frac{1}{2}(0.3)(65.319)^2-\frac{1}{2}(0.3)V_3^2 = (0.16)(0.3*9.8)(2)$

Volving for $V_3$ , we get:

$V_3 = 65.27 m/s$

3. Finally, equation 3 is equal to:

$\frac{1}{2}MV_3^2 = \frac{1}{2}K_2X_2^2$

where $K_2$ is the constant of the second spring and $X_2$ is the compress of the second spring, so:

$\frac{1}{2}(0.3)(65.27)^2 = \frac{1}{2}(2)X_2^2$

solving for $X_2$ , we get:

$X_2=25.27m$

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nikitadnepr [17]

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3 years ago

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Answer:

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The length of an arc is given as:

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3 years ago

if the vessel in the sample problem accelerates fir 1.00 min, what will its speed be after that minute ?

LUCKY_DIMON [66]

Answers:

a) 154.08 m/s=554.68 km/h

b) 108 m/s=388.8 km/h

Explanation:

<u>The complete question is written below: </u>

<u></u>

<em>In 1977 off the coast of Australia, the fastest speed by a vessel on the water was achieved. If this vessel were to undergo an average acceleration of $1.80 m/s^{2}$ , it would go from rest to its top speed in 85.6 s. </em>

<em>a) What was the speed of the vessel? </em>

<em> </em>

<em>b) If the vessel in the sample problem accelerates for 1.00 min, what will its speed be after that minute? </em>

<em></em>

<em>Calculate the answers in both meters per second and kilometers per hour</em>

<em></em>

a) The average acceleration $a_{av}$ is expressed as:

$a_{av}=\frac{\Delta V}{\Delta t}=\frac{V-V_{o}}{\Delta t}$ (1)

Where:

$a_{av}=1.80 m/s^{2}$

$\Delta V$ is the variation of velocity in a given time $\Delta t$ , which is the difference between the final velocity $V$ and the initial velocity $V_{o}=0$ (because it starts from rest).

$\Delta t=85.6 s$

Isolating $V$ from (1):

$V=a_{av}\Delta t + V_{o}$ (2)

$V=(1.80 m/s^{2})(85.6 s) + 0 m/s$ (3)

$V=154.08 \frac{m}{s}$ (4)

If $1 km=1000m$ and $1 h=3600 s$ then:

$V=154.08 \frac{m}{s}=554.68 \frac{km}{h}$ (4)

b) Now we need to find the final velocity when $\Delta t=1 min=60 s$ :

<em></em>

$V=(1.80 m/s^{2})(60 s) + 0 m/s$ (5)

$V=108 \frac{m}{s}=388.8 \frac{km}{h}$ (6)

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