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Galina-37 [17]
2 years ago
6

Help me with the following problem

Physics
1 answer:
Snezhnost [94]2 years ago
6 0

The electric field at arbitrary point outside the sphere is determined as the E = σr³/k.

<h3>Electric field determined from Gauss law</h3>

The electric field of the surface is determined from Gauss law as shown below;

E ∫ds = Q/ε

E (4πr²) = Q/ε

E = Q/4πεr² . r

E = \frac{Q R}{4\pi \varepsilon r^3}

<h3>Electric field outside the sphere with dielectric with polarization</h3>

P = \frac{\sigma}{E} \\\\E = \frac{\sigma}{P}

E = \frac{\sigma}{k/r^3} \\\\E = \frac{\sigma r^3}{k}

where;

  • σ is dipole moment of atom of the metal
  • k is dielectric constant

Thus, the electric field at arbitrary point outside the sphere is determined as the E = σr³/k.

The complete question:

A metal sphere of radius R carries a total charge Q. outside the sphere is a dielectric with polarization p(f) k/r^3er. Determine the electric field at arbitrary point outside the sphere.

Learn more about electric field here: brainly.com/question/14372859

#SPJ1

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A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule
ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

p_i = 0

The final total momentum is instead:

p_f = m_a v_a + m_c v_c

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m_a = 125 kg is the mass of the astronaut

v_a = 2.50 m/s is the velocity of the astronaut

m_c = 1900 kg is the mass of the capsule

v_c is the velocity of the capsule

Since the total momentum must be conserved, we have

p_i = p_f = 0

so

m_a v_a + m_c v_c=0

Solving the equation for v_c, we find

v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

F \Delta t = \Delta p

The change in momentum of the astronaut is

\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s

And the duration of the push is

\Delta t = 0.600 s

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

K=\frac{1}{2}mv^2

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J

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