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BlackZzzverrR [31]

3 years ago

9

If we think of the value system, as a river, and each stop along the river is a port where value was added. What was the startin

g point, the furthest upstream, in the value system in the film industry?

1 answer:

denis23 [38]3 years ago

6 0

Answer:

The raw materials.

Explanation:

The starting point of every manufactured object is the raw material. In the value system of the film industry, the starting point is the raw material which includes chemicals used in the manufacture of films and cameras.

Therefore, If we think of the value system, as a river, and each stop along the river is a port where value was added. The starting point, the furthest upstream, in the value system in the film industry is the raw materials.

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An arcade ball is thrown with an initial speed of 7.0 m/s and follows the trajectory shown. The ball enter the basket .95 second

dmitriy555 [2]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

$x = 4.70 \ m$ , $y = 0.2803 \ m$

Explanation:

From the question we are told that

The initial velocity of the arcade ball is $u = 7.0 \ m/s$

The time taken by the ball to enter the basket is $t = 0.95 \ s$

Generally the x -component of the initial velocity is

$u_x = 7 * cos (45)$

=> $u_x = 4.95 \ m/s$

Generally the y -component of the initial velocity is

$u_x = 7 * sin(45)$

=> $u_y = 4.95 \ m/s$

Generally the distance x is mathematically represented as

$x = u_x * t$

=> $x = 4.95 * 0.95$

=> $x = 4.70 \ m$

Generally from kinematic equation the distance y is mathematically represented as

$y = u_y * t - \frac{1}{2} * g * t^2$

=> $y = 4.95 * 0.95 - \frac{1}{2} * 9.8 * 0.95^2$

=> $y = 0.2803 \ m$

7 0

3 years ago

22. State any three features of the electroscope.

MatroZZZ [7]

-1- was created in the 1600 by william gilbert
-2-When the charge is positive, electrons in the metal of the electroscope are attracted to the charge and move upward out of the leaves. This results in the leaves to have a temporary positive charge and because like charges repel, the leaves separate. When the charge is removed, the electrons return to their original positions and the leaves relax
3-

An electroscope is made up of a metal detector knob on top which is connected to a pair of metal leaves hanging from the bottom of the connecting rod. When no charge is present the metals leaves hang loosely downward. But, when an object with a charge is brought near an electroscope, one of the two things can happen.

7 0

3 years ago

A car accelerates uniformly in a straight line with acceleration 10m/s 2 and travels 150m in a time interval of 5s. How far will

spayn [35]

Explanation:

Given:

Acceleration of car = 10 m/s

Distance travelled in 5 sec = 150 m

To find:

Distance travelled in the next 5 seconds

Concept:

There are 2 ways to app this kind of questions .

Either , we can find the total distance travelled in 10 seconds and then subtract 150 m from it.

Whereas , you can find out the final velocity at the end of the 5th seconds. Using equation of motion, then get the distance travelled in next 5 seconds.

Calculation:

v² = u² + 2as

=> ( u + at)² = u² + 2as

=> u² + 2uat + a²t² = u² + 2as

=> 2uat + (at)² = 2as

=> 2u (10)(5) + (10 × 5)² = 2 (10)(150)

=> 100u = 3000 - 2500

=> 100u = 500

=> u = 5 m/s

Distance travelled in 10 seconds :

s = ut + ½at²

=> s = (5 × 10) + ½(10)(10)²

=> s = 50 + 500

=> s = 550 m

Distance travelled in the 2nd half will be :

d = 550 - 150

=> d = 400 m

So final answer is :

6 0

3 years ago

A horizontal spring with stiffness 0.4 N/m has a relaxed length of 11 cm (0.11 m). A mass of 21 grams (0.021 kg) is attached and

riadik2000 [5.3K]

Answer:

0.6983 m/s

Explanation:

k = spring constant of the spring = 0.4 N/m

L₀ = Initial length = 11 cm = 0.11 m

L = Final length = 27 cm = 0.27 m

x = stretch in the spring = L - L₀ = 0.27 - 0.11 = 0.16 m

m = mass of the mass attached = 0.021 kg

v = speed of the mass

Using conservation of energy

Kinetic energy of mass = Spring potential energy

(0.5) m v² = (0.5) k x²

m v² = k x²

(0.021) v² = (0.4) (0.16)²

v = 0.6983 m/s

5 0

4 years ago

In 1976, the SR-71A, flying at 20 km altitude (T = –56 0C), set the official jet-powered aircraft speed record of 3530 km/hr (21

Lapatulllka [165]

To solve this problem we will apply the concepts related to the calculation of the speed of sound, the calculation of the Mach number and finally the calculation of the temperature at the front stagnation point. We will calculate the speed in international units as well as the temperature. With these values we will calculate the speed of the sound and the number of Mach. Finally we will calculate the temperature at the front stagnation point.

The altitude is,

$z = 20km$

And the velocity can be written as,

$V = 3530km/h (\frac{1000m}{1km})(\frac{1h}{3600s})$

$V = 980.55m/s$

From the properties of standard atmosphere at altitude z = 20km temperature is

$T = 216.66K$

$k = 1.4$

$R = 287 J/kg$

Velocity of sound at this altitude is

$a = \sqrt{kRT}$

$a = \sqrt{(1.4)(287)(216.66)}$

$a = 295.049m/s$

Then the Mach number

$Ma = \frac{V}{a}$

$Ma = \frac{980.55}{296.049}$

$Ma = 3.312$

So front stagnation temperature

$T_0 = T(1+\frac{k-1}{2}Ma^2)$

$T_0 = (216.66)(1+\frac{1.4-1}{2}*3.312^2)$

$T_0 = 689.87K$

Therefore the temperature at its front stagnation point is 689.87K

6 0

4 years ago