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trapecia [35]
3 years ago
6

A traditional unit of length in Japan is the ken (1 ken 1.97 m). What are the ratios of (a) square kens to square meters and (b)

cubic kens to cubic meters? What is the volume of a cylindrical water tank of height 5.50 ken
Physics
2 answers:
Roman55 [17]3 years ago
7 0
If 1 ken is 1.97 meter, then 1 square ken is 3.8809 square meters, and one cubic ken is 7.645373. As for the cylindrical tank, the volume of it would be 10.835 times the radius of the cylinder time 1.97^2 times pi. As you didn't specify the radius, I can't give the exact answer but that would be how to get it.
OleMash [197]3 years ago
3 0

Answer:

Part a)

A = 3.88 m^2

Part b)

V = 7.64 m^3

Explanation:

1 ken = 1.97 m

Part a)

square ken = ______square m

now we can say

A = 1 ken^2

A = 1 (1.97 m)^2

A = 3.88 m^2

Part b)

cubic Ken = _____cubic meter

now we can say

V = 1 ken^3

V = 1 (1.97 m)^3

V = 7.64 m^3

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In the Millikan oil drop experiment the measured charge of any single droplet was always a whole number multiple of -1.60 x 10-1
BaLLatris [955]

Answer:

Number of electrons, n = 6

Explanation:

Total charge in a single droplet, q=-9.6\times 10^{-19}\ C

The measured charge of any single droplet, e=-1.6\times 10^{-19}\ C

Let n is the number of excess electrons are contained within the drop. According to the quantization of charge :

q=ne

n=\dfrac{q}{e}

n=\dfrac{-9.6\times 10^{-19}}{-1.6\times 10^{-19}}

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So, there are 6 electrons contained within the drop. Hence, this is the required solution.

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3 years ago
SOH-CAH-TOA is used to solve for the ________
Misha Larkins [42]

Answer:

c. initial (x and y)

Explanation:

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Therefore, we will have to make use of trigonometric ratios which is also known by the mnemonic "SOH CAH TOA"

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3 years ago
When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a
sergij07 [2.7K]

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = \frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}

  = 4.59\times 10^{-19} \ J

or,

  = \frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}

  = 2.87 \ eV

(b)

As we know,

⇒ Vq=\frac{hc}{\lambda}-\Phi_0

By substituting the values, we get

⇒ 1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0

⇒                       \Phi_0=2.3\times 10^{-19} \ J

or,

⇒                            =\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}

⇒                            =1.4375 \ eV

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3 years ago
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Lelu [443]

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Explanation:

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A meteor is the flash of light that we see in the night sky when a small chunk of interplanetary debris burns up as it passes through our atmosphere. "Meteor" refers to the flash of light caused by the debris, not the debris itself.

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5 0
3 years ago
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