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2 years ago

What would happen if the pilot did not keep the airplane "trimmed"

Physics

1 answer:

Mademuasel [1]2 years ago

3 0

Answer:

In explanation

Explanation:

Pilots who dont use trim often like the feeling of holding constant back pressure because The heavier control forces makes it more difficult to over-control the airplane inside the turn, so it gives the sense of a more stable flight

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How might observations of the sun's outermost layers reveal what's happening in the interior? and how could this information be

LuckyWell [14K]

A raging activity can be found in the Sun's interior, with pressure waves being produced and travelling back and forth, from the core to the surface and back to the core. By looking closely at the 'surface' we can see these "ripples". It gives us an idea of how dense the material was that the waves passed through. In a way, this can help to predict solar storms in the future.

8 0

3 years ago

Read 2 more answers

The cheetah is considered the fastest running animal in the world. Cheetahs can accelerate to a speed of 21.7 m/s in 2.50 s and

viktelen [127]

Answer:

1) 64.2 mi/h

2) 3.31 seconds

3) 47.5 m

4) 5.26 seconds

Explanation:

t = Time taken = 2.5 s

u = Initial velocity = 0 m/s

v = Final velocity = 21.7 m/s

s = Displacement

a = Acceleration

1) Top speed = 28.7 m/s

1 mile = 1609.344 m

$1\ m=\frac{1}{1609.344}\ miles$

1 hour = 60×60 seconds

$1\ s=\frac{1}{3600}\ hours$

$28.7\ m/s=\frac{\frac{28.7}{1609.344}}{\frac{1}{3600}}=64.2\ mi/h$

Top speed of the cheetah is 64.2 mi/h

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow t=\frac{21.7-0}{2.5}\\\Rightarrow a=8.68\ m/s^2$

Acceleration of the cheetah is 8.68 m/s²

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{28.7-0}{8.68}\\\Rightarrow t=3.31\ s$

It takes a cheetah 3.31 seconds to reach its top speed.

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{28.7^2-0^2}{2\times 8.68}\\\Rightarrow s=47.5\ m$

It travels 47.5 m in that time

4) When s = 120 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 120=0\times t+\frac{1}{2}\times 8.68\times t^2\\\Rightarrow t=\sqrt{\frac{120\times 2}{8.68}}\\\Rightarrow t=5.26\ s$

The time it takes the cheetah to reach a rabbit is 120 m is 5.26 seconds

8 0

2 years ago

You are sitting on a merry-go-round at a distance of 2m from its center. It spins 15 times in 3 min. What distance do you move a

soldier1979 [14.2K]

Answer:

A) 12.57 m

B) 5 RPM

C) 3.142 m/s

Explanation:

A) Distance covered in 1 Revolution:

The formula that gives the relationship between the arc length or distance covered during circular motion to the angle subtended or the revolutions, is given as follows:

s = rθ

where,

s = distance covered = ?

r = radius of circle = 2 m

θ = Angle = 2π radians (For 1 complete Revolution)

Therefore,

s = (2 m)(2π radians)

s = 12.57 m

B) Angular Speed:

The formula for angular speed is given as:

ω = θ/t

where,

ω = angular speed = ?

θ = angular distance covered = 15 revolutions

t = time taken = 3 min

Therefore,

ω = 15 rev/3 min

ω = 5 RPM

C) Linear Speed:

The formula that gives the the linear speed of an object moving in a circular path is given as:

v = rω

where,

v = linear speed = ?

r = radius = 2 m

ω = Angular Speed in rad/s = (15 rev/min)(2π rad/1 rev)(1 min/60 s) = 1.571 rad/s

Therefore,

v = (2 m)(1.571 rad/s)

v = 3.142 m/s

8 0

2 years ago

What are the effects of gravity on earth?

Sliva [168]

Answer:

Without it, we could not survive on Earth. Earth orbits the sun due to the gravity of the sun, which maintains us at a convenient distance from it to enjoy the sun's warmth and light. It keeps the air we need to breathe and our atmosphere in place. Our planet is held together by gravity.

5 0

1 year ago

Read 2 more answers

A uniformly charged solid disk of radius R = 0.45 m carries a uniform charge density of σ = 175 μC/m². A point P is located a di

siniylev [52]

Answer:

1408.685 KN/C

Explanation:

Given:

R = 0.45 m

σ = 175 μC/m²

P is located a distance a = 0.75 m

k = 8.99*10^9

The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:

$E = 2*pi*k*o * (1 - \frac{a}{\sqrt{a^2 + R^2} })\\$

part a)

Electric Field strength at point P: a = 0.75 m

$E = 2*pi*8.99*10^9*175*10^-6 * (1 - \frac{0.75}{\sqrt{0.75^2 + 0.45^2} })\\\\E = 9885021.285*(0.1425070743)\\\\E = 1408.685 KN/C$

part b)

Since, R >> a, we can approximate a / R = 0 ,

Hence, E simplified relation becomes:

$E = 2*pi*k*o * (1 - \frac{a/R}{\sqrt{a^2/R^2 + 1} })\\\\E = 2*pi*k*o * (1 - \frac{0}{\sqrt{0 + 1} })\\\\E = 2*pi*k*o$

E = σ / 2*e_o

part c)

Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:

Electric Field strength due to point charge is:

E = k*δ*pi*R^2 / a^2

Since, R << a, Surface area = δ*pi

Hence,

E = (k*δ*pi/a^2)

6 0

2 years ago