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3 years ago

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What is the final speed of an object that starts from rest and accelerates uniformly at 3.0 meters per second2 over a distance o

f 6.0 meters? (1) 8.0 m/s (2) 6 m/s (3) 16 m/s (4) 64 m/s

1 answer:

jasenka [17]3 years ago

7 0

Answer:

16 m/s sir

Explanation:

because i am in collage

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The gravitational force of attraction between two students sitting at their desks in physics class is 2.59 × 10−8 N. If one stud

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<h2>The distance between students is 2.46 m</h2>

Explanation:

The force of attraction due to Newton's gravitation law is

F = $\frac{Gm_1m_2}{r^2}$

Here G is the gravitational constant

m₁ is the mass of one student

m₂ is the mass of second student .

and r is the distance between them

Thus r = $\sqrt{\frac{Gm_1m_2}{F} }$

If we substitute the values in the above equation

r = $\sqrt{\frac{6.673x10^-^1^1x31.9x30.0}{2.59x10^-^8} }$

= 2.46 m

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Yes, they are.

Explanation:

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Which of the following is not a property of displacement?

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Answer:

Your correct answer is option C. Displacement is equal to the total distance traveled by an object in motion.

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This is something that is definitely not a displacement.

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A piston-cylinder device initially contains 1.4 kg saturated liquid water at 200oC. Now heat is transferred to the water until t

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Answer:

Explanation:

Given

mass of saturated liquid water $m=1.4\ kg$

at $200^{\circ}$ specific volume is $\nu =0.001157\ m^3\kg$ (From Table A-4,Saturated water Temperature table)

$V_1=m\nu _1$

$V_1=1.4\times 0.001157$

$V_1=1.6198\times 10^{-3}\ m^3$

Final Volume $V_2=4V_1$

$V_2=4\times (1.6198\times 10^{-3})$

$V_2=6.4792\times 10^{-3}\ m^3$

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$\nu _2=\frac{V_2}{m}$

$\nu _2=\frac{6.4792\times 10^{-3}}{1.4}$

$\nu _2=0.004628\ m^3/kg$

Now we see the value and find the temperature it corresponds to specific volume at vapor stage in the table.

$T_2=T_1^{*}+\frac{T_2^{*}-T_1^{*}}{\alpha _2^{*}-\alpha _1^{*}}\times (\alpha _2-\alpha _1^{*})$

$T_2=370^{\circ}+\frac{373.95-370}{0.003106-0.004953}\times (0.004628-0.004953)$

$T_2=370.7^{\circ} C$

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