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Sladkaya [172]
2 years ago
11

Assuming a current is flowing, what increases the strength of the magnetic field of a coiled electrical wire?.

Physics
1 answer:
lesya692 [45]2 years ago
4 0

Answer:The amount of current flowing

Explanation:this derives from amperes law of maxwell’s equations. It’s easiest to see in differential form. The curl of the magnetic field is proportional to the current plus the change (time derivative) in electric field

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3 years ago
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A uniformly charged sphere has a total charge of 300uc and a radius of 8cm. Find the electric field density at A point 16cm from
s2008m [1.1K]

E = <u>kQ</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>

(r + h)²

where,

k = 9 × 10^9Nm²C^-2

Q = total charge, 300uC = 300 × 10^ -6C

r = 8 × 10^ -2m

h = 16 × 10^ -2m

then,

E = <u>9</u><u>e</u><u>9</u><u> </u><u>*</u><u> </u><u>3</u><u>0</u><u>0</u><u>e</u><u>^</u><u>-</u><u>6</u><u> </u><u> </u><u> </u><u> </u>

(8e^-2 + 16e^-2)²

E = 4687500N/C

6 0
2 years ago
Which of these processes describes the effect Earth's atmosphere has on Earth's hydrosphere?
Ede4ka [16]
Warm, moist air increasing ocean temp
6 0
3 years ago
Choose all the answers that apply. Oxygen is _____. transported by blood picked up in the alveoli released into the air by the l
aleksklad [387]
ANSWER:

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4 0
2 years ago
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The 2.50 kg cube in the figure has edge lengths d = 6.50 cm and is mounted on an axle
kozerog [31]

Answer:

0.191 s

Explanation:

The distance from the center of the cube to the upper corner is r = d/√2.

When the cube is rotated an angle θ, the spring is stretched a distance of r sin θ.  The new vertical distance from the center to the corner is r cos θ.

Sum of the torques:

∑τ = Iα

Fr cos θ = Iα

(k r sin θ) r cos θ = Iα

kr² sin θ cos θ = Iα

k (d²/2) sin θ cos θ = Iα

For a cube rotating about its center, I = ⅙ md².

k (d²/2) sin θ cos θ = ⅙ md² α

3k sin θ cos θ = mα

3/2 k sin(2θ) = mα

For small values of θ, sin θ ≈ θ.

3/2 k (2θ) = mα

α = (3k/m) θ

d²θ/dt² = (3k/m) θ

For this differential equation, the coefficient is the square of the angular frequency, ω².

ω² = 3k/m

ω = √(3k/m)

The period is:

T = 2π / ω

T = 2π √(m/(3k))

Given m = 2.50 kg and k = 900 N/m:

T = 2π √(2.50 kg / (3 × 900 N/m))

T = 0.191 s

The period is 0.191 seconds.

7 0
3 years ago
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