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3 years ago

Describe a situation when you might travel at a high velocity bit with low acceleration

1 answer:

5 0

I am sitting in my seat.
I am listening to my mp3 and reading my book.
My eyes are getting heavy. They start to close.
I try to stay awake, but it's no use.
I am so warm and comfortable and sleepy,
and I have just finished my dinner.
Finally I can't help it. Resistance is futile.
I give up, and fall deep asleep.
My head rests back against my soft, comfy seat.

My seat is in row 26 on the airplane I'm flying in
to visit my grandmother on the coast.
We are cruising at 560 miles an hour, bearing 280°,
at flight level 320 .
The temperature outside my window is -60°F .

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A 60 cm diameter wheel accelerates from rest at a rate of 7 rad/s2. After the wheel has undergone 14 rotations, what is the radi

Mamont248 [21]

Answer:

$a=368.97\ m/s^2$

Explanation:

Given that,

Initial angular velocity, $\omega=0$

Acceleration of the wheel, $\alpha =7\ rad/s^2$

Rotation, $\theta=14\ rotation=14\times 2\pi =87.96\ rad$

Let t is the time. Using second equation of kinematics can be calculated using time.

$\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s$

Let $\omega_f$ is the final angular velocity and a is the radial component of acceleration.

$\omega_f=\omega_i+\alpha t\\\\\omega_f=0+7\times 5.01\\\\\omega_f=35.07\ rad/s$

Radial component of acceleration,

$a=\omega_f^2r\\\\a=(35.07)^2\times 0.3\\\\a=368.97\ m/s^2$

So, the required acceleration on the edge of the wheel is $368.97\ m/s^2$ .

6 0

3 years ago

to measure the static friction coefficient between a block and a vertical wall, a spring is attached to the block, is pushed on

Stolb23 [73]

Answer:

μ = mg/kx

Explanation:

Since the bock does not slip, the frictional force equals the weight of the block. So, F = mg. Now, the frictional force, F = μN where μ = coefficient of static friction and N = Normal force.

Now, the normal force equals the spring force F' = kx where k = spring constant and x = compression of spring.

N = F' = kx

So, F = μN = μkx

μkx = mg

So, μ = mg/kx

8 0

3 years ago

Two pipes move the same amount of ideal fluid in the same amount of time. One pipe has a 2 in. diameter; the other has a 3 in. d

KATRIN_1 [288]

Answer:

a) 3-in. pipe

Explanation:

Given that

Fluid flow is in same amount in the same time it means that volume flow rate is same for the pipes

Volume flow rate

Q = A V

A=Area ,V=Velocity

$A=\dfrac{\pi}{4}d^2$

If diameter d is more then the velocity will be less for same volume flow rate .We also Know that if pressure is more then the velocity will be less.

The second pipe 3 in diameter having more diameter then the velocity will be less but the pressure will be more.

That is why the 3 in diameter is having more pressure than 2 in diameter pipe.

Therefore the answer will be a.

a) 3-in diameter pipe

6 0

3 years ago

a glass bottle of soda is sealed with a screw cap the absolute pressure of the gas inside the bottle is 1.80*10^5 pa. assuming t

eduard

The force exerted by a pressure of any gas over a surface its given by the formula P=F/S (where P is pressure, F force and S surface).

We can multiply both sides of the formula by S to obtain the force.

P*S=(F*S)/S

P*S=F

Solve for P=1.80*10^5 Pa and S=4.10*10^-4 m^2 ([Pa] =[N/m^s])

(1.80*10^5 N/m^s) * (4.10*10^-4 m^2) =F

73.8 N =F

8 0

3 years ago

Wrapping paper is being unwrapped from a 5.0-cm radius tube, free to rotate on its axis. if it is pulled at the constant rate of

lisov135 [29]

So the equation for angular velocity is

Omega = 2(3.14)/T

Where T is the total period in which the cylinder completes one revolution.

In order to find T, the tangential velocity is

V = 2(3.14)r/T

When calculated, I got V = 3.14

When you enter that into the angular velocity equation, you should get 2m/s

5 0

3 years ago