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xz_007 [3.2K]
3 years ago
14

In a region of space, a magnetic field points in the +x-direction (toward the right). Its magnitude varies with position accordi

ng to the formula Bx=B0+bx, where B0 and b are positive constants, for x≥0. A flat coil of area A moves with uniform speed v from right to left with the plane of its area always perpendicular to this field. Part A What is the emf induced in this coil while it is to the right of the origin?
Physics
1 answer:
otez555 [7]3 years ago
3 0

Answer:

E=Abv

Explanation:

We are given that

B_x=B_0+bx

For x\geq 0

Area of coil=A

Speed=v

We have to find the emf induced in the coil while it is to the right of the origin.

We know that induced emf

E=-\frac{d\phi}{dt}

\phi=BA

E=-\frac{d(A(B_0+bx)}{dt}

E=-A(b)\frac{dx}{dt}

We know that

v=-\frac{dx}{dt}

Substitute the value

E=Abv

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Driving force increases, friction forces increase, the driving force is bigger than friction 12.

Explanation:

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3 years ago
Which would you rather have generating electricity for your city: nuclear fission reactions (traditional nuclear power), nuclear
Arada [10]

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Fossil Combustion Reactions

Explanation:

It's more efficient (I'll edit later)

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3 years ago
An ideal spring is hung vertically from the ceiling. The spring constant is k = 125 N/m. A block of mass m = 650 g (1000 g = 1 k
Brrunno [24]

Answer:

0.102 m

Explanation:

k = spring constant of the spring = 125 N/m

m = mass of the block attached to the spring = 650 g = 0.650 kg

x = maximum extension of the spring

h = height dropped by the block = x

Using conservation of energy

Spring potential energy gained = Gravitational potential energy lost

(0.5) k x² = mgh

(0.5) k x² = mgx

(0.5) (125) x = (0.650) (9.8)

x = 0.102 m

3 0
3 years ago
Two water balloons of mass 0.75 kg collide and bounce off of each other without breaking. Before the collision, one water balloo
bagirrra123 [75]
By the law of momentum conservation:-
=>m¹u¹ + m²u² = m1v1 + m²v² {let East is +ve}
=>u¹ + u² = v¹ + v² {as m1=m2}
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5 0
3 years ago
A cruise ship sails due south at 2.00 m/s while a coast guard patrol boat heads 19.0° north of east at 5.60 m/s. What are the x-
Lilit [14]

Answer:

The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

Explanation:

Given that,

Velocity of ship = 2.00 m/s due south

Velocity of boat = 5.60 m/s due north

Angle = 19.0°

We need to calculate the component

The velocity of the ship in term x and y coordinate

v_{s_{x}}=0

v_{s_{y}}=2.0\ m/s

The velocity of the boat in term x and y coordinate

For x component,

v_{b_{x}}=v_{b}\cos\theta

Put the value into the formula

v_{b_{x}}=5.60\cos19

v_{b_{x}}=5.29\ m/s

For y component,

v_{b_{y}}=v_{b}\sin\theta

Put the value into the formula

v_{b_{y}}=5.60\sin19

v_{b_{y}}=1.82\ m/s

We need to calculate the x-component and y-component of the velocity of the cruise ship relative to the patrol boat

For x component,

v_{sb_{x}}=v_{s_{x}}-v_{b_{x}}

Put the value into the formula

v_{sb_{x}=0-5.29

v_{sb}_{x}=-5.29\ m/s

For y component,

v_{sb_{y}}=v_{s_{y}}-v_{b_{y}}

Put the value into the formula

v_{sb_{x}=2.-1.82

v_{sb}_{x}=0.18\ m/s

Hence, The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

7 0
3 years ago
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