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Elina
3 days ago
10

A physicist is constructing a solenoid. She has a roll of insulated copper wire and a power supply. She winds a single layer of

the wire on a tube with a diameter of dsolenoid = 10.0 cm. The resulting solenoid is ℓ = 60.0 cm long, and the wire has a diameter of dwire = 0.100 cm. Assume the insulation is very thin, and adjacent turns of the wire are in contact. What power (in W) must be delivered to the solenoid if it is to produce a field of 6.40 mT at its center? (The resistivity of copper is 1.70 ✕ 10−8 Ω · m.)
Physics
1 answer:
ValentinkaMS [3.3K]3 days ago
5 0
The power required is P = 105.44 W. From the provided data: Diameter D = 10 cm, Length L = 60 cm, wire diameter d = 0.1 cm, magnetic field strength B = 6.4 mT, and resistivity of copper ρ = 1.7 x 10⁻⁸ Ω·m. The turns in the solenoid N can be found using N = L/d, thus N = 60/0.1 = 600 turns. The total length of the wire is Lc = πDN, calculated as Lc = 3.14 x 0.1 x 600 equaling 188.4 m. Current I is determined as I = 5.09 A, and the resistance R is identified as R = 4.07 Ω. The power P is then calculated using the formula P = I²R, thus P = 5.09² x 4.07 W results in P = 105.44 W.
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Answer: 24.24 m

Explanation:

A player launches a football 50.0 m at an angle of 61° to the north of west. We will break this down into vertical and horizontal elements.

Horizontal component: 50 cos 61° = 24.24 m directed westward

Vertical component: 50 sin 61° = 43.73 m directed toward the north.

Refer to the diagram below.

Therefore, the westward displacement of the football corresponds to the horizontal component of the displacement, which is 24.24 m.

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The capacitors in each circuit are fully charged before the switch is closed. Rank, from longest to shortest, the length of time
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This involves circuit analysis through simplification of the resistors and capacitors. We need to determine the time constant for each circuit in figures A, B, C, D, and E. This leads to ranking the duration the bulbs remain lit from longest to shortest based on each circuit's time constant. The ranking for the time constants is C > A = E > B > D. Capacitance plays a pivotal role in electrostatics, and devices called capacitors are vital components in electronic circuits. When more charge is applied to a conductor, the voltage escalates proportionately. The capacitance of a conductor is quantified as C = q/v. Adding resistors in series raises resistance while parallel configurations reduce it, conversely increasing capacitance in parallel and diminishing it in series. Thus, circuits with greater time constants take longer to discharge.
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On its own, a certain tow-truck has a maximum acceleration of 3.0 m/s2. what would be the maximum acceleration when this truck w
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Let us denote a_1=3 m/s^2 as the highest acceleration of the truck by itself. The force generated by the engine to propel the truck in this scenario is
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F=3m a_2
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5 0
1 month ago
Read 2 more answers
Consider a 4-mg raindrop that falls from a cloud at a height of 2 km. When the raindrop reaches the ground, it won't kill you or
inna [2995]

Answer:

The work performed by air resistance totals -0.0782 J

Explanation:

Hello!

According to the principle of conservation of energy, the energy of a raindrop must remain constant.

At the outset, the raindrop possesses only gravitational potential energy:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the raindrop.

g = gravitational acceleration (9.8 m/s²)

h = height.

Let's determine the initial potential energy of the raindrop:

(4 mg should be converted into kg: 4 mg · 1 kg / 1 × 10⁶ mg = 4 × 10⁻⁶ kg)

PE = 4 × 10⁻⁶ kg · 9.8 m/s² · 2000 m

PE = 0.0784 J

As the raindrop descends, some of its potential energy converts into kinetic energy while the rest is lost to the air resistance. Upon reaching the ground, all initial potential energy has been either turned into kinetic energy or spent overcoming air resistance:

initial PE = final KE + Work by air

Where:

KE = kinetic energy.

Work by air = work done by air resistance.

The kinetic energy at ground level is computed as follows:

KE = 1/2 · m · v²

Where:

m = mass

v = velocity

<pThus:

KE = 1/2 · 4 × 10⁻⁶ kg · (10 m/s)²

KE = 2 × 10⁻⁴ J

Now, we can find the work done by air resistance:

initial PE = final KE + Work by air

0.0784 J = 2 × 10⁻⁴ J + Work by air

Work by air = 0.0784 J - 2 × 10⁻⁴ J

Work by air = 0.0782 J

Since work is performed in the opposite direction to movement, this results in a negative value. Therefore, the work done by air resistance is -0.0782 J.

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