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2 years ago

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Your employer asks you to build a 18-cmcm-long solenoid with an interior field of 5.8 mTmT . The specifications call for a singl

e layer of wire, wound with the coils as close together as possible. You have two spools of wire available. Wire with a #26 gauge is 0.41 mm in diameter and can carry up to 1 A. Which wire should you use, and what current will you need?

1 answer:

andrezito [222]2 years ago

5 0

Answer:

Explanation:

The magnetic field in a solenoid is

B = μ₀ N / L I

Where N is the number of turns, L the solenoid length and I the current

N = B L / μ₀ I

Let's calculate

N = 5.8 10⁻³ 0.18 / 4 π 10⁻⁷ 1

N = 8.3 102 laps

N = 831 laps

Let's find the solenoid length

For this we use a rule of proportions

L_solenoid = Turns * wire diameter

L_ solenoid = 831 * 0.41 10--3

L_solenoid = 0.3407 m

We see that two turns are needed in the wire to have a length of 0.18 m

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During the 440, a runner changes his speed as he comes out of the curve onto the home stretch from 18 ft/sec to 38 ft/sec over a

Sloan [31]

Answer:

$6.67ft/s^2$

Explanation:

We are given that

Initial velocity=u=18ft/s

Final velocity,v=38ft/s

Time=t=3 s

We have to find the average acceleration over that 3 s period.

We know that

Average acceleration,a= $\frac{v-u}{t}{t}$

Using the formula

Average acceleration,a= $\frac{38-18}{3}ft/s^2$

Average acceleration,a= $\frac{20}{3}ft/s^2$

Average acceleration,a= $6.67ft/s^2$

Hence, the average acceleration= $6.67ft/s^2$

5 0

2 years ago

A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm f

konstantin123 [22]

Answer:

(a) The total energy of the object at any point in its motion is 0.0416 J

(b) The amplitude of the motion is 0.0167 m

(c) The maximum speed attained by the object during its motion is 0.577 m/s

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

displacement of the toy, x = 0.012 m

speed of the toy, v = 0.4 m/s

(a) The total energy of the object at any point in its motion

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

E = 0.0416 J

(b) the amplitude of the motion

E = ¹/₂KA²

$A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m$

(c) the maximum speed attained by the object during its motion

$E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s$

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2 years ago

A 14000N car traveling at 25m/s rounds a curve of radius 200m. Find the following: a. The centripetal acceleration of the car.

tamaranim1 [39]

Answer:

Explanation:

Given

Weight of car $W=14,000\ N$

mass of car $m=\frac{14,000}{9.8}=1428.57\ N$

velocity of car $v=25\ m/s$

radius $r=200\ m$

(a)Centripetal acceleration is given by

$a_c=\frac{v^2}{r}$

$a_c=\frac{25^2}{200}$

$a_c=3.125\ m$

(b)Force that provide centripetal acceleration

$F=F_c=\frac{mv^2}{r}$

$F=\frac{1428.57\times 25^2}{200}$

$F=4464.285\ N$

(c)Friction force between car and tires is given by

$=\mu N$

where $\mu$ =coefficient of static friction

N=normal reaction

Centripetal force will balance the friction force

$F_c=F_r$

$4464.285=\mu \times 1428.57\times 9.8$

$\mu =0.318$

6 0

2 years ago

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A boy jumps into an indoor swimming pool. He notices that the water appears to get colder as he

Viefleur [7K]

Answer:

cold air is more dense than warm water so it sinks to the bottom of the pool

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2 years ago

A horizontal spring with spring constant 85 n/m extends outward from a wall just above floor level. a 3.5 kg box sliding across

Rina8888 [55]

k = spring constant of the spring = 85 N/m

m = mass of the box sliding towards the spring = 3.5 kg

v = speed of box just before colliding with the spring = ?

x = compression the spring = 6.5 cm = 6.5 cm (1 m /100 cm) = 0.065 m

the kinetic energy of box just before colliding with the spring converts into the spring energy of the spring when it is fully compressed.

Using conservation of energy

Kinetic energy of spring before collision = spring energy of spring after compression

(0.5) m v² = (0.5) k x²

m v² = k x²

inserting the values

(3.5 kg) v² = (85 N/m) (0.065 m)²

v = 0.32 m/s

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2 years ago