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2 years ago

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A 15.2 kg mass has a gravitational potential energy of -342 J. How high from the ground is it? Group of answer choices GPE canno

t be negative 2.3 m 22.5 m 530 m

1 answer:

vredina [299]2 years ago

4 0

A. The height above the ground reached by the mass is 2.3 m.

<h3>Height traveled by the mass</h3>

P.E = mgh

where;

m is mass
h is height reached by the object
g is acceleration due to gravity

h = P.E/mg

h = (342)/(15.2 x 9.8)

h = 2.3 m

Thus, the height above the ground reached by the mass is 2.3 m.

Learn more about gravitational potential energy here: brainly.com/question/1242059

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Answer:

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Explanation:

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3 years ago

What evidence did wegener have to support his theory of plate tectonics?

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3 years ago

A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium position with a downward velocity of

Rudiy27

Answer:

Explanation:

Given

mass of spring $m=100\ gm$

extension in spring $x=5\ cm$

downward velocity $v=70\ cm/s$

Position in undamped free vibration is given by

$u(t)=A\cos \omega _0t+B\sin \omega _0t$

where $\omega _0^2=\frac{k}{m}$

also $\frac{k}{m}=\frac{g}{L}$

$\omega _0^2=\frac{k}{m}=\frac{9.8}{0.05}$

$\omega _0=14$

$u(t)=A\cos(14t)+B\sin(14t)$

it is given

$u(0)=0$

$u'(0)=70\ cm/s$

substituting values we get

$A=0$

$u(t)=B\sin (14t)$

$u'(t)=14B\cos (14t)$

$70=14B$

$B=\frac{10}{2}$

$B=5$

$u(t)=5\sin (14t)$

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3 years ago

Why is honesty especially important in a scientist when the results of an experiment go against predictions❓❔

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Because other scientists and science in general rely on their collegues' research which in turn allows development of our knowledge on the given subject. The more dont-to-earth reason may be safety. If someone performs an experiment without knowledge of its true results it might result in some danger to the safety of those perforning it without knowledge of all the risks.

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3 years ago

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

postnew [5]

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)

$V_1=m\nu _1$

$V_1=1.4\times 0.001157$

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

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

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

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

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

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$T_2=370.7^{\circ} C$

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3 years ago