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Nina [5.8K]
3 years ago
10

A(n) ________ has charge but negligible mass, whereas a(n) ________ has mass but no charge.

Physics
2 answers:
Dahasolnce [82]3 years ago
7 0

Answer:

An electron has charge but negligible mass, whereas a neutron has mass but no charge.

Explanation:

Atoms are made of equal number of protons and electrons. Almost all the mass of an atom is in the nucleus of the atom. The nucleus is made of protons and neutrons (except hydrogen, which has no neutron)

The charge of an electron is -1.6 \times 10^{-19} C

The charge of a proton is +1.6 \times 10^{-19} C

The charge of a neutron is 0 C

The mass of a proton is 1.6726 \times 10^{-27} kg or 1.007276 u

The mass of a neutron is 1.6749 \times 10^{-27} kg or 1.008664 u

The mass of an electron is 9.1 \times 10^{-31} kg or 0.00054858 u

As we can see the proton is 1,838 times heavier than an electron and the neutron is 1,840 times heavier than an electron.

So the electron has a negligible mass and a negative charge but the neutron has no charge but it is heavier than electron and proton

vfiekz [6]3 years ago
6 0
Electron;Neutron is the correct answer.
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For safety reasons, a worker’s eye travel time in a certain operation must be separated from the manual elements that follow. Th
iogann1982 [59]

Answer:

The answer is 12.67 TMU

Explanation:

Recall that,

worker’s eyes travel  distance must be = 20 in.

The perpendicular distance from her eyes to the line of travel is =24 in

What is the MTM-1 normal time in TMUs that should be allowed for the eye travel element = ?

Now,

We solve for the given problem.

Eye travel is = 15.2 * T/D

=15.2 * 20 in/24 in

so,

= 12.67 TMU

Therefore, the MTM -1 of normal time that should be allowed for the eye  travel element is = 12.67 TMU

7 0
3 years ago
Describe why you weigh less on moon than on earth. Give examples
dmitriy555 [2]
Weight doesn't really mean much as it just means gravity the bigger a space object is the more force it has to pull on something since the moon is smaller than the earth then it has less gravity and then less weight on scales.
5 0
2 years ago
What would most likely happen as a result of the generator in a wind turbine breaking? The blades would not be turned. Less stea
Aleonysh [2.5K]

Answer:

What would most likely happen as a result of the generator in a wind turbine breaking?

The What would most likely happen as a result of the generator in a wind turbine breaking?

The blades would not be turned.

Less steam would be produced.

Electricity would not be generated.

Solar energy would not be absorbed.

The blades would not be turned.

Less steam would be produced.

Electricity would not be generated.

Solar energy would not be absorbed.

Explanation:

F

6 0
3 years ago
Read 2 more answers
A flea jumps straight up to a maximum height of 0.550 m . what is its initial velocity v0 as it leaves the ground?
Alexxx [7]
Since my givens are x = .550m [Vsub0] = unknown
 [Asubx] = =9.80
 
 [Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]

[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]) 

Vsubx is the final velocity, which at the max height is 0, and Xsub0 is just 0 as that's where it starts so I just plug the rest in

0^2 = [Vsub0x]^2 + 2[-9.80]*(.550)

0 = [Vsub0x]^2 -10.78

10.78 = [Vsub0x]^2

Sqrt(10.78) = 3.28 m/s 


3 0
3 years ago
A flywheel with a diameter of 1.42 m is rotating at an angular speed of 207 rev/min. (a) What is the angular speed of the flywhe
Archy [21]

Answer:

a. 21.68 rad/s b. 30.78 m/s c. 897 rev/min² d. 1085 revolutions

Explanation:

a. Its angular speed in radians per second  ω = angular speed in rev/min × 2π/60 = 207 rev/min × 2π/60 = 21.68 rad/s

b. The linear speed of a point on the flywheel is gotten from v = rω where r = radius of flywheel = 1.42 m

So, v = rω = 1.42 m × 21.68 rad/s = 30.78 m/s

c. Using α = (ω₁ - ω)/t where α = angular acceleration of flywheel, ω = initial angular speed of wheel in rev/min = 21.68 rad/s = 207 rev/min, ω₁ = final angular speed of wheel in rev/min = 1410 rev/min = 147.65 rad/s, t = time in minutes = 80.5/60 min = 1.342 min

α = (ω₁ - ω)/t

  = (1410 - 207)/(80.5/60)

  = 60(1410 - 207)/80.5

  = 60(1203)80.5

  = 896.65 rev/min² ≅ 897 rev/min²

d. Using θ = ωt + 1/2αt²

where θ = number of revolutions of flywheel. Substituting the values of the variables from above, ω = 207 rev/min, α = 896.65 rev/min² and  t = 80.5/60 min = 1.342 min

θ = ωt + 1/2αt²

  = 207 × 1.342 + 1/2 × 896.65 × 1.342²

  = 277.725 + 807.417

  = 1085.14 revolutions ≅ 1085 revolutions

5 0
3 years ago
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