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PolarNik [594]
2 years ago
9

Two objects having masses m1 and m2 are connected to each other as shown in the figure and are released from rest. There is no f

riction on the table surface or in the pulley. The masses of the pulley and the string connecting the objects are completely negligible. What must be true about the tension t in the string just after the objects are released?

Physics
2 answers:
GrogVix [38]2 years ago
5 0

Answer:

T<m2hh

Explanation:

Two objects having masses m1 and m2 are connected to each other as shown in the figure and are released from rest. There is no friction on the table surface or in the pulley. The masses of the pulley and the string connecting the objects are completely negligible. What must be true about the tension t in the string just after the objects are released?

Two masses  m1 and m2  is released and there is tension T in the string

a is the acceleration of the system

Therefore

For m1

T-m1g=m1a

T=m1(g+a)

m2

this is so because te weight is going downward wile tension in te rope is facing upward

m2g-T=m2a

T=m2(g-a)

from the equation ,

T<m2

tankabanditka [31]2 years ago
4 0

Answer:m_1g

Explanation:

Given

Two masses m_1  and m_2  is released and there is tension T in the string

Suppose a is the acceleration of the system

Therefore from Diagram

For m_1

T-m_1g=m_1a

T=m_1(g+a)------1

for m_2 body

m_2g-T=m_2a

T=m_2(g-a)-------2

From above two Equation it is said that Tension is greater than m_1g and less than m_2g

m_1g

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Answer:

P_{sol}=50.4\ mm.Hg

Explanation:

According to given:

  • molecular mass of glycerin, M_g=3\times 12+8+3\times 16=92\ g.mol^{-1}
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  • ∵Density of water is 0.992\ g.cm^{-3}= 0.992\ g.mL^{-1}
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  • pressure of mixture, P_x=55.32\ torr= 55.32\ mm.Hg
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<em>Upon the formation of solution the vapour pressure will be reduced since we have one component of solution as non-volatile.</em>

<u>moles of water in the given quantity:</u>

n_w=\frac{m_w}{M_w}

n_w=\frac{313.5}{18}

n_w=17.42 moles

<u>moles of glycerin in the given quantity:</u>

n_g=\frac{m_g}{M_g}

n_g=\frac{154}{92}

n_g=1.674 moles

<u>Now the mole fraction of water:</u>

X_w=\frac{n_w}{n_w+n_g}

X_w=\frac{17.42}{17.42+1.674}

X_w=0.912

<em>Since glycerin is non-volatile in nature so the vapor pressure of the resulting solution will be due to water only.</em>

\therefore P_{sol}=X_w\times P_x

\therefore P_{sol}=0.912\times 55.32

P_{sol}=50.4\ mm.Hg

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Power = 13.5744 kilowatts

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Power is the rate at which work is done. ... Great power means a large amount of work or energy developed in a short time. For example, when a powerful car accelerates rapidly, it does a large amount of work and consumes a large amount of fuel in a short time.

Formula for power = work/time

= IVT/T

= IV

Where I is the current

And V is the voltage

The voltage V supply = 168 v

The current A supply = 80.8 A

Power = 80.8*168

Power = 13574.4 watts

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Answer:

(a) \overrightarrow{L}=885.5\widehat{k}

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(a) Angular momentum is given by

\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}

Where, p is the linear momentum and r is the position vector about which the angular momentum is calculated.

Here, \overrightarrow{r}=3\widehat{i}-4\widehat{j}

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\overrightarrow{p}=2.3\left ( 40\widehat{i}+75\widehat{j} \right )

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So, the angular momentum

\overrightarrow{L}=\left ( 3\widehat{i}-4\widehat{j} \right )\times\left ( 92\widehat{i}+172.5\widehat{j} \right )

\overrightarrow{L}=885.5\widehat{k}

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\overrightarrow{L}=1046.5\widehat{k}

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