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ANTONII [103]
3 years ago
13

(LC)Light fixtures and placement that create shadows on the set, that obscure or completely hide action in certain areas of the

set, or that change as the main character’s emotional state changes are all ways that lighting can be used to heighten the drama and suspense in dramatic films. True/False
Physics
1 answer:
Sedaia [141]3 years ago
5 0

The correct answer is true.

It is true that light fixtures and placement that create shadows on the set, that obscure or completely hide action in certain areas of the set, or that change as the main character’s emotional state changes are all ways that lighting can be used to heighten the drama and suspense in dramatic films.

Lighting plays an important role in film making because it can create scenes that enhance the de drama of the moment or the right mood that the director wants to share. Lighting in the film is an art because the basic principle is that the scene needs to look natural. From that principle, filmmakers and light specialist cand create many kinds of dramatic or jubilation moments if they know how to apply light principles to each scene.

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The driver of a 1000 kg car traveling at a speed of 16.7 m/s applies the brakes. If the brakes provide a force of - 8000 N to st
RSB [31]

Answer:

Given:

m=1000kg

u= 16.7m/s

v=0m/s

F=8000N

Required:

s=?

Solution:

F=m × a

8000N=1000kg × a

a=8m/s^2

Since it decelerate a= -8m/s^2

v^2 = u^2 + 2as

s=v^2 - u^2 / 2a

s= 0 - (16.7m/s)^2 / 2 × -8m/s^2

s= -278.89/-16

s= 17.43m

The car travels approximately 17.43m before it stops

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6 0
2 years ago
Why a paperclip does NOT float?
Nataliya [291]
Because of Surface tension
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2 years ago
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Three objects are attached to a massless rigid rod that has an axis of rotation as shown. Assuming all of the mass of each objec
Vitek1552 [10]
The aggregate of all the given moment of inertia's will be the moment of inertia of this system.
as, moment of inertia is given as
l = m * r^2
so, finding the moment of inertia of all the individual and adding them

<span>I=2∗<span>1^2</span>+1∗<span>2^2</span>+.5∗<span>2.5^2
</span>=9.125</span>
3 0
3 years ago
The magnitude of the electrical force acting between a +2.4 × 10–8 C charge and a +1.8 × 10–6 C charge that are separated by 0.0
monitta
Use Coulomb law: F = k * q1*q2 / (r^2), where k = 9.00 * 10^9 N.m^2/C^2

F = 9.00 * 10^9 N.m^2/C^2 * 2.4*10^-8 C * 1.8*10^-6 C / [0.008m]^2 = 38.88 * 10^ -5 N

F = 39 * 10 -5N


4 0
3 years ago
Read 2 more answers
8-14 A Cu-30% Zn alloy tensile bar has a strain-hardening coefficient of 0.50. The bar, which has an initial diameter of 1 cm an
77julia77 [94]

Answer:

The true stress at true strain 0.05cm/cm is 80MPa

Explanation:

Given that

the strength coefficient is K

true strain is ε

strain hardening exponent is n

initial diameter of bar is d = 1cm, (10mm)

tensile force is F

engineering stress(S) = 120

the engineering stress(S) = \frac{F}{\frac{\pi }{4}(d^2) }

To find force (F) =

                    120 = \frac{F}{\frac{\pi }{4}(100^{2} )}

                     F = 120 * (π/4) * (100)

                     F = 9425N

Calculate the true strain  (ε) = In (l₀ / l₁)

where

l₀ =  initial length of the metallic bar = 3cm

l₁ = final length of metallic bar = 3.5cm

ε = In (3.5 / 3)

  = In 1.1667

  = 0.154cm/cm

Calculate the true stress (σ) at fracture point

          = \frac{F}{\frac{\pi }{4}(d^2) }}

tensile force is F and final diameter of bar is d₁ (d in the eqn)

Substitute 9425 N for F and 0.926 cm (9.26mm) for d₁ (d in the eqn)

σ = \frac{9425}{\frac{\pi }{4}(9.26^2) }}

   = 140MPa

To find the strength coefficient (K) of the material bar

K = \frac{140}{\sqrt{0.154} }

K = \frac{140}{0.3925}

   = 356.75MPa

To calculate the true stress σ true strain of 0.05cm/cm

K  = 356.75MPa

σ = 356.75(0.05)^0^.^5

  = 356.75 ( 0.2236)

  = 80MPa

The true stress at true strain 0.05cm/cm is 80MPa

6 0
3 years ago
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