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Bess [88]
3 years ago
8

A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In

itially the field is zero and then changes to 1.50 T, pointing upward when viewed from above, perpendicular to the circular plane, in a time of 130 ms. What is the average induced emf around the border of the circular region?
Physics
1 answer:
Daniel [21]3 years ago
3 0

Answer:81.57\mu V

Explanation:

Given

radius of circular region r=1.50 mm

A=\pi r^2=7.069\times 10^{-6}\ m^2

Magnetic Field B=1.50\ T

time t=130 ms

Flux is given by

\phi =B\cdot A

change in Flux d\phi =(B_f-B_i)A

Emf induced is e=\frac{\mathrm{d} \phi}{\mathrm{d} t}

e=\frac{(1.5)\cdot 7.069\times 10^{-6}}{130\times 10^{-3}}

e=81.57 \mu V

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What color do glacier appear if they are moving towards us?
kirill [66]

Answer:

Subcategories: Ice field

Explanation:

Glacier ice is blue because the red (long wavelengths) part of white light is absorbed by ice and the blue (short wavelengths) light is transmitted and scattered. The longer the path light travels in ice, the more blue it appears.

3 0
3 years ago
How long should a spring be stretched for it to store 45 J of energy? The force constant of the spring is
ELEN [110]

Answer:

x = 0.4 m

Explanation:

When a spring is stretched from its equilibrium position. Some energy is stored in the spring. This energy is called the elastic potential energy of the spring. The formula used to calculate the magnitude of this stored energy is given as follows:

P.E = (1/2)kx²

where,

P.E = Elastic Potential Energy Stored in the spring = 45 J

k = Spring Constant = 540 N/m

x = amount of stretching = ?

Therefore,

45 J = (1/2)(540 N/m)x²

x² = (45 J)(2)/(540 N/m)

x = √(0.167 m²)

<u>x = 0.4 m</u>

5 0
3 years ago
two barges full of salted toad guts have a collision. the red barge has a mass of 150000kg and is traveling northwest at 0.25m/s
solniwko [45]

The final velocity of the red barge in the collision elastic is 0.311 m/s when it collides with blue barge pf mass 1000000 kg.

Final velocity(v3)  of the red barge is calculated by following formula

m1×v1+ m2×v2= (m1+m2)v3

Substituting the value of m1= 150000 kg, v1= 0.25 m/s, m2= 1000000 kg, v2= 0.32 m/s

150000 × 0.25+ 1000000×0.32= (150000+1000000)×v3

37500+ 320000= 1150000×v3

357500= 1150000×v3

v3= 0.311 m/s

<h3>What is elastic collision velocity? </h3>
  • The velocity of the target particle after a head-on elastic impact in which the projectile is significantly more massive than the target will be roughly double that of the projectile, but the projectile velocity will remain virtually unaltered.

For more information on elastic collision velocity kindly visit to

brainly.com/question/29051562

#SPJ9

6 0
1 year ago
A 65-kg person stands on a scale in a moving elevator while holding a 5.0 kg mass suspended from a massless spring with spring c
disa [49]

Answer:

The question is incomplete. However, I believe, it is asking for the acceleration of the elevator. This is 3.16 m/s².

Explanation:

By Hooke's law, F = ke

F is the force on a spring, k is the spring constant and e is the extension or compression.

From the question,

F = (1.08\text{ kN/m}) \times (6.0 \times 10^{-2}\text{ m}) = 64.8 \text{ N}

This is the force on the mass suspended on the spring. Its acceleration, a, is given by

F = ma

a = \dfrac{F}{m}

a = \dfrac{64.8 \text{ N}}{5\text{ kg}} = 12.96\text{ m/s}^2

This acceleration is more than the acceleration due to gravity, g = 9.8 m/s². Hence the elevator must be moving up with an acceleration of

12.96 - 9.8 m/s² = 3.16 m/s²

7 0
3 years ago
Playing tetherball, a boy hits the 0.45 kg ball, moving toward him initially at 4.6 m/s, in the opposite direction at 2.5 m/s. H
saul85 [17]

Answer:

HI

Explanation:

8 0
3 years ago
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