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3 years ago

What is one way a C47 can be used on set?

Engineering

2 answers:

just olya [345]3 years ago

8 0

It transfers military men

S_A_V [24]3 years ago

3 0

Answer:

C47 may seem like a fancy word, but in the film world, it is the name of one of the simplest, most useful and versatile tools, which is commonly knows as a clothespin.

Following are major uses of C47 on a film set.

1) Lightning Purposes

C47 are used for attaching materials like gel and diffusion to the adjustable flaps, that control the direction of light

2) Adjusting Wardrobes

C47 are used to shorten, tighten or redesign a material of clothing for a specific purpose. It avoids permanent change and the hassle of sewing.

3) Modifications

These can be used for temporary repairs or to modify the set and shooting locations, e.g if a curtain is not hanging properly, C47 can be used

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(a) 36+45.00

(b) 24+65.00

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3 years ago

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According to the

zysi [14]

Answer:

The part of the system that is considered the resistance force is;

Explanation:

The simple machine is a system of pulley that has two pulleys

The effort, which is the input force at A gives the value of the tension at C and D which are used to lift the load B

Therefore, we have;

A = C = D

B = C + D = C + C = 2·C

∴ C = B/2

We have;

C = B/2 = A

Therefore, with the pulley only a force, A equivalent to half the weight, B, of the load is required to lift the load, B

The resistance force is the constant force in the system that that requires an input force to overcome in order for work to be done

It is the force acting to oppose the sum of the other forces system, such as a force acting in opposition to an input force

Therefore, the resistance force is the load force, B, for which the input force, A, is required in order for the load to be lifted.

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3 years ago

An overhead 25m long, uninsulated industrial steam pipe of 100mm diameter is routed through a building whose walls and air are a

DedPeter [7]

Answer:

a) he rate of heat loss from the steam line is 18.413588 kW

b) the annual cost of heat loss from line is $12904.25

Explanation:

first we find the area

A = πdL

d is the diameter (0.1m) and L is the length (25m)

A = π × 0.1 × 25

A = 7.85 m²

Now rate of heat loss through convection

qconv = hA(Ts -Ta)

h is the convective heat transfer coefficient (10 W/m²K), Ts is surface temperature (150°), Ta is temperature of air (25°)

so we substitute

qconv = 10 W/m²K × 7.85 m² × ( 150° - 25°)

qconv = 9817.477 J/s

Now heat lost through radiation

qrad = ∈Aα ( Ts⁴ - Ta⁴)

∈ is the emissivity (0.8), α is the boltzmann constant ( 5.67×10⁻⁸m⁻²K⁻⁴ ),

first we shall covert our temperatures from Celsius to kelvin scale

Ts is surface temperature (150 + 273K ), Ta is temperature of air (25 + 273K)

so we substitute

qrad = 0.8 × 7.854 × 5.67×10⁻⁸ × ( (423)⁴ - (298)⁴ )

qrad = 3.5625×10⁻⁷ × 2.413×10¹⁰

qrad = 8596.112 J/s

Now to get the total rate of heat loss through convection and radiation, we say

q = qconv + qrad

q = 9817.477 + 8596.112

q = 18413.588 J/s ≈ 18.413588 kW

Therefore the rate of heat loss from the steam line is 18.413588 kW

annual cost of heat lost rate

A = C × q/n × ( 3600 × 24 × 365 )

C is the cost of heat per MJ( $0.02/10⁶) n is broiler efficiency ( 0.9)

so we substitute

A = 0.02/10⁶ × 18413.588/0.9 × ( 3600 × 24 × 365 )

A = $12904.25

Therefore the annual cost of heat loss from line is $12904.25

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4 years ago

An object is supported by a crane through a steel cable of 0.02m diameter. If the natural swinging of the equivalent pendulum is

devlian [24]

Answer:

22.90 × 10⁸ kg

Explanation:

Given:

Diameter, d = 0.02 m

ωₙ = 0.95 rad/sec

Time period, T = 0.35 sec

Now, we know

T= $2\pi\sqrt{\frac{L}{g}}$

where, L is the length of the steel cable

g is the acceleration due to gravity

0.35= $2\pi\sqrt{\frac{L}{9.81}}$

L = 0.0304 m

Now,

The stiffness, K is given as:

K = $\frac{\textup{AE}}{\textup{L}}$

Where, A is the area

E is the elastic modulus of the steel = 2 × 10¹¹ N/m²

K = $\frac{\frac{\pi}{4}d^2\times2\times10^11}{0.0304}$

K = 20.66 × 10⁸ N

Also,

Natural frequency, ωₙ = $\sqrt{\frac{K}{m}}$

mass, m = $\sqrt{\frac{K}{\omega_n^2}}$

mass, m = $\sqrt{\frac{20.66\times10^8}{0.95^2}}$

mass, m = 22.90 × 10⁸ kg

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3 years ago

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olganol [36]

Answer:

A. Use your emergency brake

Explanation:

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Do not forget to remove the handbrake before starting. If we drive with the handbrake on, the clutch and brakes will wear quickly. Similarly, it is not advisable to use it when hydraulic brakes fail, and less if the vehicle is going at high speed.

The best way to care for the handbrake is to use it only when parking the vehicle. The handbrake has cables that allow pressure on the wheel brakes, and this helps the hydraulic brakes not wear out. In turn, vehicles with disc brakes on the rear wheels have a mechanism called a self-adjuster, which keeps the brake pads aligned avoiding wear.

If we feel any annoying sound when braking, or the braking distance is longer than usual, there is probably a problem with the brake pads. The handbrake and hydraulic brake work under different mechanisms, so whenever you feel that there is a problem with braking, you should consult the mechanic.

If the car is automatic, when is the parking brake used? In the case of cars with automatic transmission, when parking the car is at P and the transmission is immobilized. If the vehicle is on a flat surface, this part exerts little pressure, but if it is on a slope, the entire weight of the vehicle rests on the brake and the transmission itself, therefore the parking brake must be applied.

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3 years ago

Read 2 more answers