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3 years ago

6

The refrigeration cycle uses _____.

2 answers:

makkiz [27]3 years ago

7 0

the answer is A. the laws of thermodynamics

Stolb23 [73]3 years ago

4 0

A. The laws of thermodynamics

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A student is performing a double-slit experiment to determine the wavelength of a light source. She has measured the distance be

Harman [31]

Answer:

lights need elewctricity

Explanation:

4 0

3 years ago

A simple pendulum consists of a mass M attached to a string oflength L andnegligible mass. For this system, when undergoing smal

PilotLPTM [1.2K]

The frequency of the pendulum is independent of the mass on the end. (c)

This means that it doesn't matter if you hang a piece of spaghetti or a school bus from the bottom end. If there is no air resistance, and no friction at the top end, and the string has no mass, then the time it takes the pendulum to swing from one side to the other <u><em>only</em></u> depends on the <u><em>length</em></u> of the string.

8 0

2 years ago

2in+3in+4in+4in+6in+7in calculate the area and perimeter of each shape

ololo11 [35]

Perimeter=26in

Area would be to multiply the units together

4 0

2 years ago

Read 2 more answers

A 2.10 cm × 2.10 cm square loop of wire with resistance 1.30×10−2 Ω has one edge parallel to a long straight wire. The near edge

Troyanec [42]

Answer:

$I_{l} =44.84 \mu A$

Explanation:

given,

side of square loop = a = 2.10 cm

Resistance of the wire = 1.30×10⁻² Ω

Length of the loop = c = 1.10 cm

rate of increasing current = 130 A/s

$\phi = \dfrac{\mu_0Ib}{2\pi}(ln(\dfrac{c+a}{c}))$

$\dfrac{d\phi}{dt}= \dfrac{\mu_0b}{2\pi}\dfrac{dI}{dt}(ln(\dfrac{c+a}{c}))$

$I_{l} = \dfrac{V}{R}$

$I_{l} = \dfrac{1}{R}\dfrac{d\phi}{dt}$

$I_{l} = \dfrac{1}{R}\dfrac{\mu_0b}{2\pi}\dfrac{dI}{dt}(ln(\dfrac{c+a}{c}))$

$I_{l} = \dfrac{1}{1.3 \times 10^{-2}}\dfrac{4\pi\times 10^{-7}\times 0.021}{2\pi}\times 130\times (ln(\dfrac{0.011+0.021}{0.011}))$

$I_{l} =44.84 \times 10^{-6}\A$

$I_{l} =44.84 \mu A$

3 0

3 years ago

A blue-green photon (λ = 488 nm ) is absorbed by a free hydrogen atom, initially at rest. What is the recoil speed of the hydrog

Natalka [10]

Answer:

The recoil speed is $2.207\times 10^{4} m/s$

Solution:

Wavelength of a blue-green photon, $\lambda_{BG} = 488 nm = 488\times 10^{- 9} m$

Now, the energy associated with the blue-green photon:

$E_{BG} = \frac{hc}{\lambda_{BG}}$

where

h = Planck's constant

C = speed of light ion vacuum

$E_{BG} = \frac{6.626\times 10^{- 34}\times 3\times 10^{8}}{488\times 10^{- 9}}$

$E_{BG} = 4.07\times 10^{- 19} J$

Also, we know that the recoil speed can be calculated by the KInetic energy which is equal to the Energy of the blue-green photon:

$KE_{H} =\frac{1}{2}m_{p}v_{H}$

where

$v_{H}$ = velocity of Hydrogen atom

$m_{p} = 1.67\times 10^{- 27} kg$ = mass of H-atom

Now,

$KE_{H} =\frac{1}{2}m_{p}(v_{H})^{2}$

$4.07\times 10^{- 19} =\frac{1}{2}\times 1.67\times 10^{- 27}\times (v_{H})^{2}$

$v_{H} = \sqrt(4.87\times 10^{8}) = 2.207\times 10^{4} m/s$

7 0

2 years ago