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Marrrta [24]
3 years ago
12

Lamar writes several equations trying to better understand potential energy. What conclusion is best supported by Lamar’s work?

The elastic potential energy is the same for any distance from a reference point. The gravitational potential energy equals the work needed to lift the object. The gravitational potential energy is the same for any distance from a reference point. The elastic potential energy equals the work needed to stretch the object.
Physics
2 answers:
faust18 [17]3 years ago
7 0
Its B i got it right on the test
ehidna [41]3 years ago
3 0
<h2>Answer:</h2>

The correct answer is B.

B. The gravitational potential energy equals the work needed to lift the object.

<h2>Explanation:</h2>

Lamar writes several equations trying to best understand potential energy. The conclusion of his work is "The gravitational potential energy equals the work needed to lift the object".

<h2 />
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Discuss the effects of breaking a food chain.
Tom [10]

Explanation:

Food chain is linear sequence of the organisms through which the nutrients and the energy pass as one organism eats the another.

<u>If one species in the food chain ceases to exist, the other members in rest of chain could cease to exist too or may increase in population which can affect other food chains. (Breaking of food chain).</u>

This is because the animals in the food chain are dependent on the nutrient present in the another.

For example in the food chain,

Grass ----  Deer ------ Lion

If all deer extinct, the lion will die as they will have nothing to feed on and the population of grass increases.

8 0
3 years ago
I need help with this
fredd [130]
We have here what is known as parallel combination of resistors.

Using the relation:

\frac{1}{ r_{eff} } = \frac{1}{ r_{1} } + \frac{1}{ r_{2} } + \frac{1}{ r_{3} }.. . + \frac{1}{ r_{n} } \\
And then we can turn take the inverse to get the effective resistance.

Where r is the magnitude of the resistance offered by each resistor.

In this case we have,
(every term has an mho in the end)
\frac{1}{10000} + \frac{1}{2000} + \frac{1}{1000} \\ \\ = \frac{1}{1000} ( \frac{1}{10} + \frac{1}{2} + \frac{1}{1} ) \\ \\ = \frac{1}{1000} ( \frac{31}{20}) \\ \\ = \frac{31}{20000}

To ger effective resistance take the inverse:
we get,
\frac{20000}{31} \: ohm \\ = 645 .16 \: ohm

The potential difference is of 9V.

So the current flowing using ohm's law,

V = IR

will be, 0.0139 Amperes.
7 0
3 years ago
Imagine a large block that sinks in a tub of water. You hold a much smaller block of the exact same material in your hand, above
goldenfox [79]
It will sink because mass does not affect the physical properties of the object.
3 0
3 years ago
Gravitational potential energy is related to–A. what an object is made of. B.an object’s motion.C. an object’s position.D. an ob
Bas_tet [7]

Answer:

The Answer is B because the material the object is made of, the position, or the color have absolutely nothing to do with gravitational potential energy

7 0
2 years ago
Se lanza verticalmente una esfera con una rapidez de 30m/se. Determinar la rapidez de la esfera a una altura de 40m (g=10m/s2)
sergeinik [125]

v^2-{v_0}^2=2a(x-x_0)

dónde v es la velocidad de la esfera, v_0 es suya velocidad inicial, a=-g la aceleración debida a la gravedad, x la posición, y x_0 la posición inicial. Tomamos x_0=0\,\mathrm m a referirse a la posición de la esfera en el momento que la esfera fue lanzada.

Entonces

v^2-\left(30\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-10\,\dfrac{\mathrm m}{\mathrm s^2}\right)(40\,\mathrm m)

\implies v^2=100\,\dfrac{\mathrm m^2}{\mathrm s^2}\implies v=\pm10\,\dfrac{\mathrm m}{\mathrm s}

Esto nos dice que la esfera alcanza una altura de 40 m en dos momentos - una vez hacia arriba y una vez hacia abajo. Sin embargo, independientemente del signo de la velocidad, sabemos que suya magnitud es 10 m/s, y así tenemos una rapidez de 10 m/s también en ambos momentos.

4 0
3 years ago
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