Fill in the missing words thats all but I don't know so yeah

From whom should a departing VFR aircraft request radar traffic information during ground operations?

zavuch27 [327]

Answer:From whom should a departing VFR aircraft request radar traffic information during ground operations? ... Answer: Sequencing to the primary Class C airport, traffic advisories, conflict resolution, and safety alerts.

Explanation:

6 0

3 years ago

A 1 900-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.00 m before coming into contact

Leno4ka [110]

Answer:

471392.4 N

Explanation:

From the question,

Just before contact with the beam,

mgh = Fd.................... Equation 1

Where m = mass of the beam, g = acceleration due to gravity, h = height. F = average Force on the beam, d = distance.

make f the subject of the equation

F = mgh/d................ Equation 2

Given: m = 1900 kg, h = 4 m, d = 15.8 = 0.158 m

Constant: g = 9.8 m/s²

Substitute into equation 2

F = 1900(4)(9.8)/0.158

F = 471392.4 N

6 0

3 years ago

Water flows through a water hose at a rate of Q1 = 860 cm3/s, the diameter of the hose is d1 = 1.85 cm. A nozzle is attached to

Snowcat [4.5K]

Answer:

a) A1 = $\frac{\pi (d1)^{2} }{4}$

b) A1 = 2.688 cm $^{2}$

c) Q1 = A1 x v1

d) v1 = 3.1994 m/s

e) A2 = $\frac{A1 X v1}{v2}$

f) A2 = 0.7963cm $^{2}$

Explanation:

a) Area = $\pi r^{2}$

r = $\frac{d}{2}$

thus,

area = $\pi (\frac{d}{2})^{2}$

A1 = $\frac{\pi (d1)^{2} }{4}[/tex]b) d1 = 1.85 cmsubstituting in the above equation,A1 = [tex]\frac{\pi (d1)^{2} }{4}$

A1 = $\frac{\pi (1.85)^{2} }{4}$

A1 = 2.688 cm $^{2}$

c) Flow rate = Area x velocity ( refer brainly.com/question/13997998)

Q1 = A1 x v1

d) From the above equation,

v1 = $\frac{Q1}{A1}$ = $\frac{860}{2.688}$ = 319.94 cm/s = 3.1994 m/s

e) Since the flow rate Q1 is constant throughout the hose, Av is a constant.

i.e. A1 x v1 = A2 x v2

thus,

A2 = $\frac{A1 X v1}{v2}$

f) v2 = 10.8 m/s.

substituting the values in the above equation,

A2 = $\frac{2.688 X 3.1994}{10.8}$ = 0.7963cm $^{2}$

3 0

4 years ago

English schoolteacher who proposed the atomic theory model of matter

vodka [1.7K]

Answer:

Ernest Rutherford.

Explanation:

Ernest Rutherford

4 0

3 years ago

A massless string connects a 1.00 kg mass to a 3.00 kg cart which is resting on a frictionless horizontal surface. The mass hang

Romashka-Z-Leto [24]

Both masses will have the same acceleration. The cart accelerates to the right with a magnitude of 4.9 m/ $s^{2}$ . The correct answer is 4.90 m/ $s^{2}$

Given that a massless string connects a 1.00 kg mass to a 3.00 kg cart which is resting on a frictionless horizontal surface.

Let M = 1kg and m = 3 kg

Since the horizontal surface is frictionless, the tension in the string will be the same. when the mass is hanged over a frictionless pulley, the tension will also be the same.

When the mass is released, the cart accelerates to the right can be calculated from Newton' second law of motion. That is,

M( g + a) = m(g - a)

1(9.8 + a) = 3( 9.8 - a)

9.8 +a = 29.4 - 3a

collect the like terms

4a = 19.6

a = 19.6/4

a = 4.9 m/ $s^{2}$

Therefore, the cart accelerates to the right with a magnitude of 4.9 m/ $s^{2}$ . The correct answer is 4.90 m/ $s^{2}$

Learn more about dynamics here: brainly.com/question/24994188

5 0

3 years ago

Fill in the missing words thats all but I don't know so yeah ​

Fill in the missing words thats all but I don't know so yeah