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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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A proton accelerates from rest in a uniform electric field of 610 N/C. At some later time, its speed is 1.4 106 m/s.(a) Find the

nata0808 [166]

Answer:

a) 5.851× 10¹⁰m/s²

b) 2.411×10⁻¹¹s

c) 1.70×10⁻¹¹m

d) 1.661×10⁻²⁷KJ

Explanation:

A proton in the field experience a downward force of magnitude,

F = eE. The force of gravity on the proton will be negligible compared to the electric force

F = eE

a= eE/m

= 1.602×10⁻¹⁹ × 610/1.67×10⁻²⁷

= 5.851× 10¹⁰m/s²

V = u + at

u= 0

v= 1.4106m/s

v= (0)t + at

t= v/a

= 1.4106m/s/5.851 ×10¹⁰

= 2.411×10⁻¹¹s

S = ut + at²

= (o)t + 5.851×10¹⁰×(2.411×10⁻¹¹)²

= 1.70×10⁻¹¹m

Ke = 1/2mv²

= (1.67×10⁻²⁷×)(1.4106)²/2

= 1.661×10⁻²⁷KJ

5 0

3 years ago

Describe how sound waves are produced

Delvig [45]

Answer:

Sound waves are produced when something vibrates.

Explanation:

The vibrating body causes the medium (water, air, etc.) Vibrations in air are called traveling longitudinal waves, which we can hear. Sound waves consist of areas of high and low pressure called compressions and rarefactions, respectively.

Sorry if this if wrong

4 0

3 years ago

Which statement correctly describes the relationship between current, voltage, and resistance? If we

Salsk061 [2.6K]

Answer: Option (b) is the correct answer.

Explanation:

According to ohm's law, the relationship between voltage, resistance, and current is that current passing through a conductor is directly proportional to the voltage over resistance.

Mathematically, I = $\frac{V}{R}$

From this relationship we can see that when we decrease the voltage, and do not change the resistance, the current will also decrease. As current is directly proportional to voltage and inversely proportional to resistance.

6 0

3 years ago

Read 2 more answers

Choose either eclipses or lunar phases and make a model of the event. Your model can take any form, providing it can explain how

babymother [125]

Answer: See the explanation below.

Explanation: For this assignment, I chose to display how eclipses are created.

My model was made utilizing a 3D displaying device program for all intents and purposes. The items utilized are three models I made for this presentation, Earth, the moon, and the sun. These three models will be utilized for the showcase.

The light that shines from the sun would create a shadow on the moon. The moon would then catch the light that should've arrived on Earth, making the shadow we call an eclipse. Earth gets a shadow of the moon and the remainder of Earth is lit up from the rest of the light, making an eclipse.

The individual I demonstrated my project to was [<em>Someone you know</em>], [<em>Pronoun</em>] said it precisely took after the occasion of an eclipse. The light from the sun being shined on to the moon rather than the Earth, creating the shadow we call an eclipse.

5 0

3 years ago

1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x

allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$

With $(\epsilon_{0})$ been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$

Explanation:

(A) Considering a uniform linear density $λ_{0}$ on the ring, then:

$dQ=\lambda dl$ (1)⇒ $Q=\lambda_{0} 2\pi R$ (2)⇒ $\lambda_{0}=\frac{Q}{2\pi R}$ (3)

Applying the technique of charge integration for finite charges:

$V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ$ (4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

$r'=\sqrt{R^{2} +Z^{2}}$ (5)

Using the expressions (1),(4) and (5) you obtain:

$V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi$

Integrating results:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$ (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

$|E| =-\nabla V$ (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

$|E| =-\frac{dV(z)}{dz}$ (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$ (S_b)

4 0

3 years ago