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3 years ago

The box set was an important because:

Physics

2 answers:

inysia [295]3 years ago

6 0

Answer:

Answer is A. It supported a more realistic acting style.

Refer below.

Explanation:

The box set was an important because:

It supported a more realistic acting style

zepelin [54]3 years ago

4 0

Answer: A. It supported a more realistic acting style

Explanation:

Box sets create the illusion of an interior room on the stage and are contrasted with earlier forms of sets which contained sliding flaps and gaps between set pieces.

In the play style of realism, the box set of the stage was a room with either a plain black backdrop or three walls. The fourth wall was invisible, separating the characters from the audience, and the ceiling was tilted down at the far end of the stage and up toward the audience. Doors slammed instead of swinging when being shut, as in reality.

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Explanation:

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Rather than ascribing the increased kinetic energy of the stone to the work of gravity, we now (when using potential energy rath

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6th grade science !!!!!!!!!!!!!!!

posledela

Answer:

Explanation:

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A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t

Mumz [18]

Answer:

$v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

$y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t$

Second stone:

$y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

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