Safety devices used in electric circuit

charle [14.2K]

Answer:

rgvdghrryyrfgtr was the other monkey that I have to do with my best pet peeve for the rest and neon scooter is my friend is giving her the money for your cuz to be a good offer tho it was a great idea for a while and then she said yes and she is a great person ever and she was madly happy I have to go soon to be honest with me if she wants me back to my base or not but she wasnt it to be honest with her but

8 0

3 years ago

Calculate the energy released by the electron-capture decay of 5727Co. Consider only the energy of the nuclei (ignore the energy

erma4kov [3.2K]

Answer:

Explanation:

⁵⁷Co₂₇ + e⁻¹ = ²⁷Fe₂₆

mass defect = 56.936296 + .00055 - 56.935399

= .001447 u

equivalent energy

= 931.5 x .001447 MeV

= 1.3479 MeV .

= 1.35 MeV

energy of gamma ray photons = .14 + .017

= .157 MeV .

Rest of the energy goes to neutrino .

energy going to neutrino .

= 1.35 - .157

= 1.193 MeV.

5 0

3 years ago

What happens to the kinetic energy of a body when: a) the mass of the body is doubled at constant velocity? b) the velocity of t

blagie [28]

Using the formula KE=1/2mv^2

a: The kinetic energy doubles.
b: The kinetic energy quadruples.
c: The kinetic energy is cut in half.
Hopefully it’s clear how the formula can show you this.

3 0

4 years ago

4. How does nearness to water affect the climate zone of an area?

Afina-wow [57]

Answer:

Choice A.

Nearness to a body of water causes an increase in humidity, due to the higher rate of evaporation.

6 0

2 years ago

Read 2 more answers

A singly charged positive ion has a mass of 3.46 × 10−26 kg. After being accelerated through a potential difference of 215 V the

jasenka [17]

Answer:

1.8 cm

Explanation:

$m$ = mass of the singly charged positive ion = 3.46 x 10⁻²⁶ kg

$q$ = charge on the singly charged positive ion = 1.6 x 10⁻¹⁹ C

$\Delta V$ =Potential difference through which the ion is accelerated = 215 V

$v$ = Speed of the ion

Using conservation of energy

Kinetic energy gained by ion = Electric potential energy lost

$(0.5) m v^{2} = q \Delta V\\(0.5) (3.46\times10^{-26}) v^{2} = (1.6\times10^{-19}) (215)\\(1.73\times10^{-26}) v^{2} = 344\times10^{-19}\\v = 4.5\times10^{4} ms^{-1}$

$r$ = Radius of the path followed by ion

$B$ = Magnitude of magnetic field = 0.522 T

the magnetic force on the ion provides the necessary centripetal force, hence

$qvB = \frac{mv^{2} }{r} \\qB = \frac{mv}{r}\\r =\frac{mv}{qB}\\r =\frac{(3.46\times10^{-26})(4.5\times10^{4})}{(1.6\times10^{-19})(0.522)}\\r = 0.018 m \\r = 1.8 cm$

5 0

3 years ago