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3 years ago

12

A superconducting solenoid has 3300 turns per meter and can carry a maximum current of 4.1 ka. find the magnetic field strength

in the solenoid.

2 answers:

jeyben [28]3 years ago

8 0

B = μ x n x I = 1.25 x 10^-6 x 3300 x 4.1 x 10^3
= 16.91 T

avanturin [10]3 years ago

7 0

Hope this helps you.

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Paladinen [302]

Answer:

Percentage of mass converted to energy = 99.9 %

Explanation:

By far most of the solar system's mass is in the Sun itself: somewhere between 99.8 and 99.9 percent. The rest is split between the planets and their satellites, and the comets and asteroids and the dust and gas surrounding our star. Seen from afar (on the scale of distances between stars) the presence of the solar system would not be obvious. We would simply see a normal-looking star. Perhaps we would pick up the presence of Jupiter, which makes up two thirds or so of the solar system outside of the Sun, by mass.

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3 years ago

A bullet of mass 0.010 kg and speed of 200 m/s is brought to rest in a wooden block after penetrating a distance of 0.10 m. The

irina [24]

Answer:

W = 200 J

Explanation:

Work will be equal to the change in kinetic energy

W = ½mv² - 0

W = ½(0.010)200²

W = 200 J

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3 years ago

Convert 9.75 millimeters to centimeters.

Alex777 [14]

0.975 is correct, to get the answer you divide the length value by 10.

4 0

3 years ago

Read 2 more answers

Use the work—energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resist

andreyandreev [35.5K]

Answer:

a) It is moving at $43.15\frac{m}{s^{2}}$ when reaches the ground.

b) It is moving at $101.44\frac{m}{s^{2}}$ when reaches the ground.

Explanation:

Work energy theorem states that the total work on a body is equal its change in kinetic energy, this is:

$W=K_f-K_i$ (1)

with W the total work, Ki the initial kinetic energy and Kf the final kinetic energy. Kinetic energy is defined as:

$K=\frac{mv^2}{2}$ (2)

with m the mass and v the velocity.

Using (2) on (1):

$W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$ (3)

In both cases the total work while the objects are in the air is the work gravity field does on them. Work is force times the displacement, so in our case is weight (w=mg) of the object times displacement (d):

$W=Fd=wd=mgd$ (4)

Using (4) on (3):

$mgd=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$ (5)

That's the equation we're going to use on a) and b).

a) Because the branch started form rest initial velocity (vi) is equal zero, using this and solving (5) for final velocity:

$v_f=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*95}$

$v_f=43.15\frac{m}{s^{2}}$

b) In this case the final velocity of the boulder is instantly zero when it reaches its maximum height, another important thing to note is that in this case work is negative because weight is opposing boulder movement, so we should use -mgd:

$-mgd=-\frac{mv_i^2}{2}$

Solving for initial velocity (when the boulder left the volcano):

$v_i=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*525}$

$v_i=101.44 \frac{m}{s^{2}}$

3 0

3 years ago

A single-turn current loop, carrying a current of 4.03 A, is in the shape of a right triangle with sides 68.1, 151, and 166 cm.

SashulF [63]

Answer:

Part a)

$F = 0$

Part b)

$F = 0.25 N$

Part c)

$F = 0.25 N$

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

$F_{net} = 0$

Explanation:

As we know that force on a current carrying wire is given as

$\vec F = i(\vec L \times \vec B)$

now we have

Part a)

current in side 166 cm and magnetic field is parallel

so we have

$F = i(\vec L \times \vec B)$

here we know that L and B is parallel to each other so

$F = 0$

Part b)

For 68.1 cm length wire we have

$F = iLB sin\theta$

here we know that

$cos\theta = \frac{68.1}{166}$

$\theta = 65.8$

so we have

$F = (4.03)(0.681)(99.3 \times 10^{-3})sin65.8$

$F = 0.25 N$

Part c)

For 151 cm length wire we have

$F = iLB sin\phi$

here we know that

$cos\phi = \frac{151}{166}$

$\theta = 24.5$

so we have

$F = (4.03)(1.51)(99.3 \times 10^{-3})sin24.5$

$F = 0.25 N$

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

$F_{net} = 0$

4 0

2 years ago