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3 years ago

A 0.200 m wire is moved parallel to a 0.500 T

Physics

1 answer:

goldenfox [79]3 years ago

8 0

Answer:

The required emf moved across the wire is zero

Explanation:

For a moving charge particle, the magnetic force can be determined by using the formula;

$\varepsilon = Bvlsin \theta$

since the wire moves in parallel, the angle $\theta$ between magnetic field and velocity = 0°

B = 0.500 T

v = 1.50 m/s

l = 0.200 m

∴

$\varepsilon = (0.500 \ T )(1.50 \ m/s) \times (0.200 \ m)\times sin (0)$

$\varepsilon = 0.15\times sin (0)$

$\varepsilon = 0$

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Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

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3 years ago

A hollow spherical shell has mass 8.20 kg and radius 0.220 m. It is initially at rest and then rotates about a stationary axis t

Likurg_2 [28]

Answer:

8.91 J

Explanation:

mass, m = 8.20 kg

radius, r = 0.22 m

Moment of inertia of the shell, I = 2/3 mr^2

= 2/3 x 8.2 x 0.22 x 0.22 = 0.265 kgm^2

n = 6 revolutions

Angular displacement, θ = 6 x 2 x π = 37.68 rad

angular acceleration, α = 0.890 rad/s^2

initial angular velocity, ωo = 0 rad/s

Let the final angular velocity is ω.

Use third equation of motion

ω² = ωo² + 2αθ

ω² = 0 + 2 x 0.890 x 37.68

ω = 8.2 rad/s

Kinetic energy,

$K = \frac{1}{2}I\omega ^{2}$

K = 0.5 x 0.265 x 8.2 x 8.2

K = 8.91 J

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3 years ago

A team of dogs accelerates a 290kg dogsled from 0 to 6.0m/s in 3.0 s. Assume that the acceleration is constant.Part AWhat is the

Mademuasel [1]

Answer:

(a) $a=2m/sec^2$

(b) 5220 j

(d) 3446.66 watt

Explanation:

We have given mass m = 290 kg

Initial velocity u = 0 m/sec

Final velocity v = 6 m/sec

Time t = 3 sec

From first equation of motion

v = u+at

So $a=\frac{v-u}{t}=\frac{6-0}{3}=2m/sec^2$

(a) We know that force is given by

F = ma

So force will be $F=290\times 2=580N$

(b) From second equation of motion we know that

$s=ut+\frac{1}{2}at^2=0\times 3+\frac{1}{2}\times 2\times 3^2=9m$

We know that work done is given by

W = F s = 580×9 =5220 j

We know that power is given as

$P=\frac{W}{t}=\frac{5220}{3}=1740Watt$

(d) Time t = 1.5 sec

So $P=\frac{W}{t}=\frac{5220}{1.5}=3466.66Watt$

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3 years ago

A ball is thrown upward.

valentina_108 [34]

Answer:

It is 10.75

Explanation:

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3 years ago

What affect does friction have on motion

satela [25.4K]

Answer: friction reduces the speed during motion

Explanation:

The more the friction, the lesser the speed during motion

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