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3 years ago

A scientific idea becomes valid when it is

Physics

1 answer:

SIZIF [17.4K]3 years ago

7 0

Answer:

are adopted when they usefully describe the world. evidence

Explanation:

A scientific idea is validated when it is published in the peer-reviewed literature in the field, has stood up to further tests, and has been positively cited.

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A pure jet engine propels an aircraft at 240 m/s through air at 45 kPa and −13°C. The inlet diameter of this engine is 1.6 m, th

erica [24]

QUESTION: A pure jet engine propels and aircraft at 340 m/s through air at 45 kPa and -13C. The inlet diameter of this engine is 1.6 m, the compressor pressure ratio is 13, and the temperature at the turbine inlet is 557C. Determine the velocity at the exit of this engines nozzle and the thrust produced.

ANSWER: Due to the propulsion from the inlet diameter of this engine bring 1.6 m allows the compressor rations to radiate allowing thrust propultion above all velocitic rebisomes.

6 0

3 years ago

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

5 0

3 years ago

Two equipotential surfaces surround a +3.10 x 10-8-c point charge. how far is the 290-v surface from the 41.0-v surface?

MrMuchimi

T<span>he equation to be used here to determine the distance between two equipotential points is:
V = k * Q / r

where v is the voltage of the point, k is a constant, Q is charge of the point measured in coloumbs and r is the distance.
In this case, we can use ratio of proportions to determine the distance between the two points. in this respect,

Point 1:
V = k * Q / r = 290
r = k*Q/290 ; kQ = 290r

Point 2:
V = k * Q / R = 41
R = k*Q/41
from equation 10 kQ = 290r
R = 290/(41)= 7.07 m
The distance between the two points then is equal to 7.07 m.

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8 0

3 years ago

Mandy is testing an unknown solution to determine whether it is an acid or a base. She places a piece of red litmus paper into t

Marina CMI [18]

The solution is a base

5 0

3 years ago

Which is the best definition of a parallel

Oksanka [162]

A parallel circuit exists when an electric charge flows in more than one path best describes it.

<h3>What is a Parallel circuit?</h3>

This type of circuit has branches in which the current divides and only part of it flows through any of the branch.

Parallel circuit having more than one branch therefore means that electric charge will flow in more than one path thereby making option A the most appropriate choice.

Read more about Parallel circuit here brainly.com/question/12069231

5 0

2 years ago