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2 years ago

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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]2 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]2 years ago

7 0

Hope this helps you.

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A spring driven dart gun propels a 10g dart. It is cocked by exerting a force of 20N over a distance of 5cm. With what speed wil

adelina 88 [10]

<span>14 m/s Assuming that all of the energy stored in the spring is transferred to dart, we have 2 equations to take into consideration. 1. How much energy is stored in the spring? 2. How fast will the dart travel with that amount of energy. As for the energy stored, that's a simple matter of multiplication. So: 20 N * 0.05 m = 1 Nm = 1 J For the second part, the energy of a moving object is expressed as KE = 0.5 mv^2 where KE = Kinetic energy m = mass v = velocity Since we now know the energy (in Joules) and mass of the dart, we can substitute the known values and solve for v. So KE = 0.5 mv^2 1 J = 0.5 0.010 kg * v^2 1 kg*m^2/s^2 = 0.005 kg * v^2 200 m^2/s^2 = v^2 14.14213562 m/s = v So the dart will have a velocity of 14 m/s after rounding to 2 significant figures.</span>

6 0

2 years ago

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The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles

belka [17]

Answer:

$F_a=5.67\times 10^{-5}\ N$

<u /> $F_b=3.49\times 10^{-5}\ N$

$F_c=9.16\times 10^{-5}\ N$

Explanation:

Given:

mass of particle A, $m_a=363\ kg$

mass of particle B, $m_b=517\ kg$

mass of particle C, $m_c=154\ kg$

All the three particles lie on a straight line.

Distance between particle A and B, $x_{ab}=0.5\ m$

Distance between particle B and C, $x_{bc}=0.25\ m$

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

<u>Force on particle A due to particles B & C:</u>

$F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}$

$F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})$

$F_a=5.67\times 10^{-5}\ N$

<u>Force on particle C due to particles B & A:</u>

<u /> $F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}$ <u />

$F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )$

$F_c=9.16\times 10^{-5}\ N$

<u>Force on particle B due to particles C & A:</u>

<u /> $F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}$ <u />

<u /> $F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2} )$ <u />

<u /> $F_b=3.49\times 10^{-5}\ N$ <u />

3 0

2 years ago

A 10 kg ball moving at 13 m/s strikes a 20 kg ball at rest. after the collision the 10 kg ball is moving with a velocity of 7m/s

Lostsunrise [7]

Answer:

30 (kg)

Explanation:

therefore the mass of the ball is 2 so 30 (kg)

6 0

2 years ago

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A large mass collides with a stationary, smaller mass. How will the masses behave if the collision is inelastic?

Mamont248 [21]

Most likely they would stick together and keep moving together

7 0

2 years ago

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For a magnetic field strength of 2T, estimate the magnitude of the maximum force on a 1-mm-long segment of a single cylindrical

nydimaria [60]

Answer:

Incomplete question

This is the complete question

For a magnetic field strength of 2 T, estimate the magnitude of the maximum force on a 1-mm-long segment of a single cylindrical nerve that has a diameter of 1.5 mm. Assume that the entire nerve carries a current due to an applied voltage of 100 mV (that of a typical action potential). The resistivity of the nerve is 0.6ohms meter

Explanation:

Given the magnetic field

B=2T

Lenght of rod is 1mm

L=1/1000=0.001m

Diameter of rod=1.5mm

d=1.5/1000=0.0015m

Radius is given as

r=d/2=0.0015/2

r=0.00075m

Area of the circle is πr²

A=π×0.00075²

A=1.77×10^-6m²

Given that the voltage applied is 100mV

V=0.1V

Given that resistive is 0.6 Ωm

We can calculate the resistance of the cylinder by using

R= ρl/A

R=0.6×0.001/1.77×10^-6

R=339.4Ω

Then the current can be calculated, using ohms law

V=iR

i=V/R

i=0.1/339.4

i=2.95×10^-4 A

i=29.5 mA

The force in a magnetic field of a wire is given as

B=μoI/2πR

Where

μo is a constant and its value is

μo=4π×10^-7 Tm/A

Then,

B=4π×10^-7×2.95×10^-4/(2π×0.00075)

B=8.43×10^-8 T

Then, the force is given as

F=iLB

Since B=2T

F=iL(2B)

F=2.95×10^-4×2×8.34×10^-8

F=4.97×10^-11N

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2 years ago