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3 years ago

Why are electrons, rather than protons, the principal charge carriers in metal wires?

1 answer:

6 0

because in any atom the electrons are in the outer orbitals while protons are within the nucleus together with the neutrons. If energy is supplied electrons can jump to higher energy levels and leave the lower orbitals.
Especially in metals the conduction band is partially filled at room temperature with allows free flow of electrons throughout the metal thus they carry charge.
(it requires huge amounts of energy to remove a proton from the nucleus such things happen on the surface of sun).

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What speed must an electron have if its momentum is to be the same as that of an x-ray photon with a wavelength of 0.30?

lorasvet [3.4K]

The momentum of the x-ray photon is p = h/lambda . Lambda is the wavelength (0.30nm=3x10^(-9)m) and h is Planck's constant,(h=6.62607004 × 10-34 m2 kg / s).The momentum is: 2.2 x 10^(-25).

The momentum can be calculated also as: p=mv, where m is the mass of the electron and v is the speed.

So v=p/m,p is known,and also the mass of the electron (m=9.10938356 × 10-31 kilograms).

v=2.2 x 10^(-25)/9.10938356 × 10-31 kilograms=0.24 x 10^6 m/s

8 0

3 years ago

Determine the minimum work per unit of heat transfer from the source reservoir that is required to drive a heat pump with therma

Sloan [31]

Answer:

The minimum work per unit heat transfer will be 0.15.

Explanation:

We know the for a heat pump the coefficient of performance ( $C_{HP}$ ) is given by

$C_{HP} = \dfrac{Q_{H}}{W_{in}}$

where, $Q_{H}$ is the magnitude of heat transfer between cyclic device and high-temperature medium at temperature $T_{H}$ and $W_{in}$ is the required input and is given by $W_{in} = Q_{H} - Q_{L}$ , $Q_{L}$ being magnitude of heat transfer between cyclic device and low-temperature $T_{L}$ . Therefore, from above equation we can write,

$&& \dfrac{Q_{H}}{W_{in}} = \dfrac{Q_{H}}{Q_{H} - Q_{L}} = \dfrac{1}{1 - \dfrac{Q_{L}}{Q_{H}}} = \dfrac{1}{1 - \dfrac{T_{L}}{T_{H}}}$

Given, $T_{L} = 460 K$ and $T_{H} = 540 K$ . So, the minimum work per unit heat transfer is given by

$\dfrac{W_{in}}{Q_{H}} = \dfrac{T_{H} - T_{L}}{T_{H}} = \dfrac{540 - 460}{540} = 0.15$

8 0

3 years ago

What is the mass of a falling rock if it produces a force of 50 N?

Rasek [7]

×× $F=m \times a$ ×

50N is your force and the acceleration is -9.8m/s^2 due to gravity.

So, you just plug that in.

$50 N=m\times-9.8m/s^2\\
\frac{50}{-9.8}=m\\
m=-5.102$

BUT you know that mass cannot be negative, so you just disregard the negative sign and the mass of the rock is 5.102 grams.

8 0

3 years ago

A car drives 40 km due east and then 50 km due west .what is the cars overall displacement?

Firdavs [7]

Answer:10 km westExplanation:he go 40 east then 50 west 50-40 is 10 so he displaces 10 km and as west is more than east in terms of km so we will say that it's 10 km west pls mark as brainliest thanks

7 0

3 years ago

Read 2 more answers

A man pushes down on a lever with 45 N and the lever moves 1.5 meters. The other end of the lever lifts a 87.6 N box .4 meters.

AysviL [449]

The input work = Force x distance
So I/P = 45 x 1.5 = 67.5 N.m

The output work = Force x distance
So, O/P = 87.6 x 0.4 = 35.04 N.m

Efficiency = (output work / Inputwork) x 100
= ( 35.04/67.5) x100

Efficiency = 51.91 %

4 0

3 years ago