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3 years ago

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3. J.J. Thomson discovered the electron in 1897. In 1904, he proposed a model

1 answer:

OLga [1]3 years ago

7 0

Answer:

Maybe this will help?

Explanation:

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What dictates the minimum and maximum pressure allowed for plumbing fixtures?

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The answer is: the building codes, which are a set of rules that regulate the conditions that a building must meet. Those requirements are very important to guarantee safety of both the people who are building it and the people who are going to work or live inside of the building in the future

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The hubbles telescopes orbit is 5.6 x10 ^5 meters above earths suface. the telescope has a mass os 1.1 x10^4 kilograms. earth ex

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3 years ago

Two bodies of specific heats S1 and S2 having the same heat capacities are combined to form a single composite body. What is the

Dafna11 [192]

$\qquad\qquad\huge\underline{{\sf Answer}}♨$

Heat capacity of body 1 :

$\qquad \sf \dashrightarrow \:m_1s_1$

Heat capacity of body 2 :

$\qquad \sf \dashrightarrow \:m_2s_2$

it's given that, the the head capacities of both the objects are equal. I.e

$\qquad \sf \dashrightarrow \:m_1s_1 = m_2s_2$

$\qquad \sf \dashrightarrow \:m_1 = \dfrac{m_2s_2}{s_1}$

Now, consider specific heat of composite body be s'

According to given relation :

$\qquad \sf \dashrightarrow \:(m_1 + m_2) s' = m_1s_1 + m_2s_2$

$\qquad \sf \dashrightarrow \:s' = \dfrac{ m_1s_1 + m_2s_2}{m_1 + m_2}$

$\qquad \sf \dashrightarrow \:s' = \dfrac{ m_2s_2+ m_2s_2}{ \frac{m_2s_2}{s_1} + m_2 }$

[ since, $m_2s_2 = m_1s_1$ ]

$\qquad \sf \dashrightarrow \:s' = \dfrac{ 2m_2s_2}{ m_2(\frac{s_2}{s_1} + 1)}$

$\qquad \sf \dashrightarrow \:s' = \dfrac{ 2 \cancel{m_2}s_2}{ \cancel{m_2}(\frac{s_2}{s_1} + 1)}$

$\qquad \sf \dashrightarrow \:s' = \dfrac{ 2 s_2}{ (\frac{s_2 + s_1}{s_1} )}$

$\qquad \sf \dashrightarrow \: s' = \dfrac{2s_1s_2}{s_1 + s_2}$

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3 years ago

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Natasha_Volkova [10]

Answer:

i think it might be D or C

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3 years ago

Diagnostic ultrasound of frequency 3.82 MHz is used to examine tumors in soft tissue. (a) What is the wavelength in air of such

maxonik [38]

Explanation:

It is given that,

Frequency of diagnostic ultrasound, f = 3.82 MHz = 3820 Hz

The speed of the sound in air, v = 343 m/s

(a) We need to find the wavelength in air of such a sound wave. Let it is given by λ₁

i.e. $\lambda=\dfrac{v}{\nu}$

$\lambda_1=\dfrac{343\ m/s}{3820\ Hz}$

$\lambda_1=0.089\ m$

(b) If the speed of sound in tissue is 1650 m/s .

$\lambda_2=\dfrac{v}{\nu}$

$\lambda_2=\dfrac{1650\ m/s}{3820\ Hz}$

$\lambda_2=0.43\ m$

Hence, this is the required solution.

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3 years ago