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olga nikolaevna [1]
3 years ago
14

The resistivity of gold is 2.44 × 10-8 Ω · m at room temperature. A gold wire that is 1.8 mm in diameter and 11 cm long carries

a current of 170 mA. How much power is dissipated in the wire?
Physics
1 answer:
tester [92]3 years ago
6 0
Power in a wire where current is flowing can be calculated from the product of the square of the current and the resistance. Resistance is equal to the product of resistivity and length divided by the area of the wire. We do as follows:

Resistance  =  2.44 × 10-8 ( 0.11) / (π)(0.0009)^2 = 1.055x10^-3 <span>Ω

P = I^2R = .170^2 (</span>1.055x10^-3 ) = 3.048x10^-5 W
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FIGURE 1 shows part of a mass spectrometer. The whole arrangement is in a vacuum. Negative ions of mass 2.84 x 10-20 kg and char
yuradex [85]

Yes, the ions can exit slit P without being deflected, if the electric field strength is 170.6 N/C

Explanation:

When the ions are inside the container, they are subjected to two forces, with directions opposite to each other:

  • The force due to the electric field, whose magnitude is F_E=qE, where q is the charge of the ion and E is the strength of the electric field
  • The force due to the magnetic field, whose magnitude is F_B=qvB, where v is the speed of the ions and B is the strength of the magnetic field

The ions will move straight and undeflected if the two forces are equal and opposite. By using Fleming Left Hand rule, we notice that the magnetic force on the (negative) ions point upward: this means that the electric field must be also upward (so that the electric force on the ions is downward). Then, the two forces are balanced if

F_E = F_B

which translates into

qE=qvB\\\rightarrow v = \frac{E}{B}

Therefore, if the speed of the ions is equal to this ratio, the ions will go undeflected.

We can even calculate the value of E at which this occurs. In fact, we know that the ions are earlier accelerated by a potential difference V=-3000 V, so we have that their kinetic energy is given by the change in electric potential energy:

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

where

q=-2.0\cdot 10^{-19}C\\m=2.84\cdot 10^{-20}kg

Solving for v, the speed,

v=\sqrt{\frac{2qV}{m}}=\sqrt{\frac{2(-2.0\cdot 10^{-19})(-3000)}{2.84\cdot 10^{-20}}}=205.6 m/s

And since the magnetic field strength is

B = 0.83 T

The strength of the electric field must be

E=vB=(205.6 m/s)(0.83 T)=170.6 N/C

Learn more about electric and magnetic fields:

brainly.com/question/8960054

brainly.com/question/4273177

brainly.com/question/3874443

brainly.com/question/4240735

#LearnwithBrainly

7 0
3 years ago
32. (FR) A mass moving at 25 m/s slides along a rough horizontal surface. The coefficient of friction is 0.30. (A) Use force and
scoray [572]

Answer:

A)s = 104.16 m

b)s= 104.16 m

Explanation:

Given that

u = 25 m/s

μ = 0.3

The friction force will act opposite to the direction of motion.

Fr= μ m g

Fr= -  m a

a=acceleration

μ m g = - m a

a= - μ g

a= - 0.3 x 10 m/s²          ( take g= 10 m/s²)

a= - 3  m/s²

The final speed of the mass is zero ,v= 0

We know that

v² = u² +2 a s

s=distance

0² = 25² - 2 x 3 x s

625 = 6 s

s = 104.16 m

By using energy conservation

Work done by all the forces =Change in the kinetic energy

- Fr.s=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2

Negative sign because force act opposite to the displacement.

- \mu\ m\ g \ s=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2

-\mu\ g \ s=\dfrac{1}{2}v^2-\dfrac{1}{2}u^2

-0.3\times 10\times \ s=\dfrac{1}{2}\times 0^2-\dfrac{1}{2}\times 25^2

- 3 x 2 x s = - 625

s= 104.16 m

4 0
3 years ago
A planet is discovered orbiting the star 51 Peg with a period of four days (0.01 years). 51 Peg has the same mass as the Sun. Me
artcher [175]

Answer:

Less than Mercury's

Explanation:

According to third Kepler's law, the square of the planet's orbital period is proportional to the cube of the average orbital radius of the planet's orbit. The constant of proportionality depends only on the mass of the star, recall that 51 Peg has the same mass as the Sun. Since the orbital period of this planet is less than Mercury's, its average orbital radius is less than Mercury's.

4 0
3 years ago
The number of electrons in a copper penny is approximately 10*10^23. How large would the force be on an object if it carried thi
slega [8]
The charge on the electron is 1.6x10^-19C. So, 10^24 of them will be a  charge of 1.6x10^5C, F = q1xq2/[(4pi epsilon nought)r^2]
3 0
3 years ago
Read 2 more answers
Starting at 9a.m., you ride your hover board for 3hrs at an average speed of 6 mph. Out of breath, you stop for tea from noon un
Elena-2011 [213]
3 times 6= 18. The average speed is 19 mph.

hope this helps!
6 0
2 years ago
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