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Ber [7]
3 years ago
6

A coil of 160 turns and area 0.20 m2 is placed with its axis parallel to a magnetic field of initial magnitude 0.40 T. The magne

tic field changes uniformly from 0.40 T in the +x direction to 0.40 T in the -x direction in 2.0 s. If the resistance of the coil is 16 Ω, at what rate is power generated in the coil?
Physics
1 answer:
ser-zykov [4K]3 years ago
8 0

Answer:

The rate at which power is generated in the coil is 10.24 Watts

Explanation:

Given;

number of turns of the coil, N = 160

area of the coil, A = 0.2 m²

magnitude of the magnetic field, B = 0.4 T

time for field change = 2 s

resistance of the coil, R =  16 Ω

The induced emf in the coil is calculated as;

emf = dΦ/dt

where;

Φ is magnetic flux = BA

emf = N (BA/dt)

emf = 160 (0.4T x 0.2 m²)/dt

emf = 12.8 V/s

The rate power is generated in the coil is calculated as;

P = V²/ R

P = (12.8²) / 16

P = 10.24 Watts

Therefore, the rate at which power is generated in the coil is 10.24 Watts

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So, the distance

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From equations (1) and (3)

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5 0
2 years ago
Which of the following are single-displacement reactions?
ololo11 [35]

Answer:

A & D

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A single-displacement reaction is a chemical reaction whereby one element is substituted for another one in a compound and thereby generating a new element and also a new compound as products.

From the options, only options A & D fits this definition of single-displacement reactions.

For option D: Both left and hand and right hand sides each have one element and one compound. We can see that K is substituted from KBr to join Cl to form KCl and Br2 on the right hand side.

For option A: Both left and hand and right hand sides each have one element and one compound. We can see that OH is substituted from 2H2O to join Mg to form Mg(OH)2 and H2 on the right hand side.

The other options are not correct because they don't involve only and element and a compound on each side of the reaction.

6 0
3 years ago
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A thin cylindrical shell and a solid cylinder have the same mass and radius. The two are released side by side and roll down, wi
Vitek1552 [10]

Answer:

Explanation:

Let the velocity be v

Total energy at the bottom

= rotational + linear kinetic energy

= 1/2 Iω² + 1/2 mv² ( I moment of inertia of shell  = mr² )

= 1/2 mr²ω² + 1/2 mv² ( v = ω r )

= 1/2 mv² +1/2 mv²

= mv²

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v² = gh

v = √gh

= √9.8 x 1.8

= 4.2 m /s

8 0
3 years ago
What ocean depth would the volume of an aluminium sphere be reduced by 0.10%
yKpoI14uk [10]

Answer:

6400 m

Explanation:

You need to use the bulk modulus, K:

K = ρ dP/dρ

where ρ is density and P is pressure

Since ρ is changing by very little, we can say:

K ≈ ρ ΔP/Δρ

Therefore, solving for ΔP:

ΔP = K Δρ / ρ

We can calculate K from Young's modulus (E) and Poisson's ratio (ν):

K = E / (3 (1 - 2ν))

Substituting:

ΔP = E / (3 (1 - 2ν)) (Δρ / ρ)

Before compression:

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After compression:

ρ+Δρ = m / (V - 0.001 V)

ρ+Δρ = m / (0.999 V)

ρ+Δρ = ρ / 0.999

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ΔP = 69×10⁹ Pa / (3 (1 - 2×0.32)) (0.001/0.999)

ΔP = 64.0×10⁶ Pa

If we assume seawater density is constant at 1027 kg/m³, then:

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