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ryzh [129]
3 years ago
14

What current is needed to generate the magnetic field strength of 5.0×10−5T at a point 1.5 cm from a long, straight wire? Expres

s your answer to two significant figures and include the appropriate units.
Physics
1 answer:
mixer [17]3 years ago
4 0

Answer:

3.7 A

Explanation:

Parameters given:

Magnetic field strength, B = 5 * 10^(-5) T

Distance of magnetic field from wire, r = 1.5 cm = 0.015 m

The magnetic field, B, due to a current, I, flowing a wire is given as:

B = (μ₀*I) / 2πr

Where μ₀ = permeability of free space

To get the current, I, we make I the subject of the formula:

I = (2πr * B) / μ₀

I = (2 * 3.142 * 5 * 10^(-5)) / (1.25663706 × 10^(-6))

I = 3.7 A

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Answer:

option (b) 4900 N

Explanation:

m = 2000 kg, R = 6380 km = 6380 x 10^3 m, Me = 5.98 x 10^24 kg, h = R

F = G Me x m / (R + h)^2

F = G Me x m / 2R^2

F = 6.67 x 10^-11 x 5.98 x 10^24 x 2000 / (2 x 6380 x 10^3)^2

F = 4900 N

4 0
3 years ago
Con lắc lò xo có độ cứng k = 100N/m được gắn vật có khối lượng m=0.1kg, kéo vật ra khỏi vị trí cân bằng 1 đoạn 5cm rồi buông tay
Delvig [45]

Answer:

The maximum velocity is 1.58 m/s.

Explanation:

A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.

Spring constant, K = 100 N/m

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w = \sqrt{K}{m}\\\\w = \sqrt{100}{0.1}\\\\w = 31.6 rad/s

The maximum velocity is

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8 0
3 years ago
Organisms that live on or burrow into the ocean floor are
creativ13 [48]
Benthos 

Option b is the answer



8 0
3 years ago
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A student attempted to measure the specific latent heat of vaporisation of water.
tensa zangetsu [6.8K]

Answer:

The latent heat of vaporization of water is 2.4 kJ/g

Explanation:

The given readings are;

The first (mass) balance reading (of the water) in grams, m₁ = 581 g

The second (mass) balance reading (of the water) in grams, m₂ = 526 g

The first joulemeter reading in kilojoules (kJ), Q₁ = 195 kJ

The second joulemeter reading in kilojoules (kJ), Q₂ = 327 kJ

The latent heat of vaporization = The heat required to evaporate a given mass water at constant temperature

Based on the measurements, we have;

The latent heat of vaporization = ΔQ/Δm

∴ The latent heat of vaporization of water = (327 kJ - 195 kJ)/(581 g - 526 g) = 2.4 kJ/g

The latent heat of vaporization of water = 2.4 kJ/g

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