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Vsevolod [243]
3 years ago
9

A 8.0-cm long solenoid has 2000 turns of wire and carries a current of 5.0-A. Calculate the strength of the magnetic field at th

e center of the solenoid in teslas.
Physics
1 answer:
CaHeK987 [17]3 years ago
6 0

Answer:

B = 0.157 T

Explanation:

Given that,

Length of the solenoid, l = 8 cm = 0.08 m

Number of turns in the wire, N = 2000

Current, I = 5 A

We need to find the strength of the magnetic field at the center of the solenoid. It is given by the formula as follows :

N=\mu_o nI, N is number of turns per unit length of solenoid.

So,

B=4\pi \times 10^{-7}\times \dfrac{2000}{0.08}\times 5\\\\=0.157\ T

So, the magnetic field at the center of the solenoid is 0.157 T.

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What is the equivalent resistance between points A and C if R1=1430, R2=1350, R3=1100, R4=1350, and R5=1150.
Marianna [84]

R1 + R4 = 1430 + 1350 = 2780 = R14    series combination of R1 & R4

R2 + R5 = 1350 + 1150 = 2500 = R25

The circuit has been reduced to 3 resistors in parallel

R314 = 2780 * 1100 / (2780 + 1100) = 788  this is the resistance of the parallel combination of R14 and R3

R31425 = 2500 * 788 / (2500 + 788) = 599 which is the equivalent of the circuit  - you can also use the formula for 3 resistors in parallel but this seems simpler

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3 years ago
A horizontal spring with stiffness 0.4 N/m has a relaxed length of 11 cm (0.11 m). A mass of 21 grams (0.021 kg) is attached and
riadik2000 [5.3K]

Answer:

0.6983 m/s

Explanation:

k = spring constant of the spring = 0.4 N/m

L₀ = Initial length = 11 cm = 0.11 m

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x = stretch in the spring = L - L₀ = 0.27 - 0.11 = 0.16 m

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v = speed of the mass

Using conservation of energy

Kinetic energy of mass = Spring potential energy

(0.5) m v² = (0.5) k x²

m v² = k x²

(0.021) v² = (0.4) (0.16)²

v = 0.6983 m/s

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