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3 years ago

Why do we need to put Airplane mode?

Physics

2 answers:

Serga [27]3 years ago

6 0

Answer:

if we ever ride a airplane we dont mess up its signals and crash ,and its easier to ignore calls and texts

Explanation:

garri49 [273]3 years ago

5 0

Assuming you mean why we need to use airplane mode when flying...

Airplane mode stops all pathways of connection coming and going from your device. It is used on planes so that the pilots can have a clear and direct connection to the ATC Tower, where workers can see the planes radar.

Hope this helps :)

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A private aviation helicopter's main rotor blades rotate at approximately

Arisa [49]

Answer: 7.5 rev/s

Explanation:

We are given the angular velocity $\omega$ a helicopter's main rotor blades:

$\omega=450 rpm=450 \frac{rev}{min}$

However, we are asked to express this $\omega$ in the International Systrm (SI) units. In this sense, the SI unit for time is second ( $s$ ):

$\omega=450 \frac{rev}{min} \frac{1 min}{60 s}$

$\omega=7.5 \frac{rev}{s}$

4 0

3 years ago

What is one joule work

natta225 [31]

It's a Newton Meter Those two are multiplied to get a joule.

8 0

3 years ago

You observe lighting striking and then hear to sound 8 seconds later. The speed of sound in air is 340 m/s. How

Leto [7]

Answer:

2.72 Kilometers

Explanation:

8 × 340 m/s = 2720 m = 2.72 Kilometers

7 0

3 years ago

A professor sits at rest on a stool that can rotate without friction. The rotational inertia of the professor-stool system is 4.

Anestetic [448]

Answer:

$\omega=0.37 [rad/s]$

Explanation:

We can use the conservation of the angular momentum.

$L=mvR$

$I\omega=mvR$

Now the Inertia is I(professor_stool) plus mR², that is the momentum inertia of a hoop about central axis.

So we will have:

$(I_{proffesor - stool}+mR^{2})\omega=mvR$

Now, we just need to solve it for ω.

$\omega=\frac{mvR}{I_{proffesor-stool}+mR^{2}}$

$\omega=\frac{1.5*2.7*0.4}{4.1+1.5*0.4^{2}}$

$\omega=0.37 [rad/s]$

I hope it helps you!

5 0

3 years ago

Calculate the potential energy of a 5 kg object sitting at the top of a 2 meter ramp.

trapecia [35]

The formula used to find potential energy is P.E. = M * G * H (P.E. is potential energy, M is mass, G is gravitational pull, and H is height). So the answer to your question is 5 * 9.8 * 2, which equals 98.

5 0

3 years ago