Using numbers to describe kinematic quantities is called what description?

An aircraft is in level flight at 225 km/hr through air at standard conditions. The lift coefficient at this speed is 0.45 and t

mojhsa [17]

Answer:

- the effective lift area for the aircraft is 8.30 m²

- the required engine thrust is 1275 N

- required power is 79.7 kW

Explanation:

Given the data in the question;

Speed V = 225 km/hr = 62.5 m/s

The lift coefficient CL = 0.45

drag coefficient CD = 0.065

mass = 900 kg

g = 9.81 m/s²

a) the effective lift area for the aircraft

we know that for a steady level flight, weight = lift and thrust = drag

Using the equation for the lift force

F $_L$ = C $_L$ $\frac{1}{2}$ ρV²A = W

we substitute

0.45 × $\frac{1}{2}$ × 1.21 × ( 62.5 )² × A = ( 900 × 9.81 )

1081.05 × A = 8829

A = 8829 / 1081.05

A = 8.30 m²

Therefore, the effective lift area for the aircraft is 8.30 m²

b) the required engine thrust and power to maintain level flight.

we use the expression for drag force

F $_D$ = T = C $_D$ $\frac{1}{2}$ ρV²A

we substitute

= 0.065 × $\frac{1}{2}$ × 1.21 × ( 62.5 )² × 8.30

T = 1275 N

Since drag and thrust force are the same,

Therefore, the required engine thrust is 1275 N

Power required;

P = TV

p = 1275 × 62.5

p = 79687.5 W

p = ( 79687.5 / 1000 )kW

p = 79.7 kW

Therefore, required power is 79.7 kW

8 0

3 years ago

580 nm light shines on a double slit

Helga [31]

Answer:

0.532

Explanation:

Your equation to find the second bright interference maximum is gonna be this: d sin (Θ) = m λ

First, find your variables.

λ = 580 · 10^-9

d = 0.000125

m = 2

Next, fill in the equation.

d sin (θ) = m λ

(0.000125) sin (θ) = (2) (580·10^-9)

Then isolate your variable.

θ = arcsin ( (2)(580·10^-9) / (0.000125) )

Run your equation and you will end up with 0.53171246 , which rounds to 0.532.

The main thing you have to watch out for is make sure you are calculating for the bright interference and not the dark interference, as well as checking you're calculating for the maximum, not the minimum.

I hope this helps :D

5 0

3 years ago

Write a description of how carbon enters and exits a consumer, such as the deer shown in the diagram.

dedylja [7]

Answer:

Carbon moves from the atmosphere to plants. In the atmosphere, carbon is attached to oxygen in a gas called carbon dioxide (CO2). Through the process of photosynthesis, carbon dioxide is pulled from the air to produce food made from carbon for plant growth.

Carbon moves from plants to animals. Through food chains, the carbon that is in plants moves to the animals that eat them. Animals that eat other animals get the carbon from their food too.

Carbon moves from plants and animals to soils. When plants and animals die, their bodies, wood and leaves decays bringing the carbon into the ground. Some is buried and will become fossil fuels in millions and millions of years.

Carbon moves from

living things to the atmosphere. Each time you exhale, you are releasing carbon dioxide gas (CO2) into the atmosphere. Animals and plants need to get rid of carbon dioxide gas through a process called respiration. Carbon moves from fossil fuels to the atmosphere when fuels are burned. When humans burn fossil fuels to power factories, power plants, cars and trucks, most of the carbon quickly enters the atmosphere as carbon dioxide gas. Each year, five and a half billion tons of carbon is released by burning fossil fuels. Of this massive amount, 3.3 billion tons stays in the atmosphere. Most of the remainder becomes dissolved in seawater.

3 0

3 years ago

A flag bearer holds a 20 N flag at the end of a pole that has an angle of 20 degrees to the horizontal. What is the force from t

Nezavi [6.7K]

I think the answer is C) i am sorry if i am wrong plz let me know if i am wrong thank u.

4 0

3 years ago

A +3.4 x 10-6 C test charge experiences forces from two other nearby charges: a 3 N force due east and a 15 N force due west. Wh

Alecsey [184]

Answer:

3.53×10⁶ N/c due west

Explanation:

From the question

E = F'/q........................ Equation 1

Where E = Electric Field, F = Net Force, q = Charge.

But,

F' = F₂-F₁...................... Equation 2

Substitute equation 2 into equation 1

E = (F₂-F₁)/q................ Equation 3

Given: F₁ = 3 N due east, F₂ = 15 N due west, q = 3.4×10⁻⁶ C

Substitute these values into equation 1

E = (15-3)/(3.4×10⁻⁶)

E = 12/(3.4×10⁻⁶)

E = 3.53×10⁶ N/c due west

4 0

3 years ago