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3 years ago

6

A 10kg crate sits on a distant planet. The planet has the same mass as Earth, but 80% of the radius of Earth (Earth radius = 637

0km.) What is the approximate weight of the object on the distant planet?
A) 120N
B) 160N
C) 80N
D) 100N

1 answer:

Nataliya [291]3 years ago

8 0

The answer is d! Hope I helped you

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Which of the following safety devices provides a weak link in a circuit that will melt if it received too much current, thus bre

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The correct answer is B. Fuse.

Fuses were commonly used in houses up to recent ages, when circuit breakers became the more popular options, as they don't have to be changed every time they melt, since they don't melt.

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a banana boat accelerates from 4.167 m/s ar 2.00m/s^2.How far has it traveled when it reaches 8.333 m/s

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Answer: The distance that we'll be travelled is d=16.35m

Explanation: The main idea here is to use the equation of motion. We are given the acceleration of 2m/s^2,the initial and final velocies. The formula to use is v^2=u^2+2as. Now we have to substitute the values.

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4 years ago

The n = 2 to n = 6 transition in the bohr hydrogen atom corresponds to the ________ of a photon with a wavelength of ________ nm

Colt1911 [192]

The energy levels of the hydrogen atom are quantized and their energy is given by the approximated formula
$E=- \frac{13.6}{n^2} [eV]$
where n is the number of the level.

In the transition from n=2 to n=6, the variation of energy is
$\Delta E=E(n=6)-E(n=2)=-13.6 ( \frac{1}{6^2}- \frac{1}{2^2} )[eV]=3.02 eV$
Since this variation is positive, it means that the system has gained energy, so it must have absorbed a photon.

The energy of photon absorbed is equal to this $\Delta E$ . Converting it into Joule,
$\Delta E=3.02 eV=4.84 \cdot 10^{-19}J$
The energy of the photon is
$E=hf$
where h is the Planck constant while f is its frequency. Writing $\Delta E=hf$ , we can write the frequency f of the photon:
$f= \frac{\Delta E}{h}= \frac{4.84 \cdot 10^{-19}J}{6.63 \cdot 10^{-34}m^2 kg/s}=7.29 \cdot 10^{14}Hz$

The photon travels at the speed of light, $c=3 \cdot 10^8 m/s$ , so its wavelength is
$\lambda = \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{7.29 \cdot 10^{14}Hz}=4.11 \cdot 10^{-7}m=411 nm$

So, the initial sentence can be completed as:
The n = 2 to n = 6 transition in the bohr hydrogen atom corresponds to the "absorption" of a photon with a wavelength of "411" nm.

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4 years ago

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Analyze the given velocity time graph

a_sh-v [17]

Answer:

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3 years ago

An infinite plane of charge has surface charge density 7.2 Î¼C/m^2. How far apart are the equipotential surfaces whose potential

Rina8888 [55]

Answer:

so the distance between two points are

$d = 0.246 \times 10^{-3} m$

Explanation:

Surface charge density of the charged plane is given as

$\sigma = 7.2 \mu C/m^2$

now we have electric field due to charged planed is given as

$E = \frac{\sigma}{2\epsilon_0}$

now we have

$E = \frac{7.2 \times 10^{-6}}{2(8.85 \times 10^{-12})}$

$E = 4.07 \times 10^5 N/C$

now for the potential difference of 100 Volts we can have the relation as

$E.d = \Delta V$

$4.07 \times 10^5 (d) = 100$

$d = \frac{100}{4.07 \times 10^5}$

$d = 0.246 \times 10^{-3} m$

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