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Leno4ka [110]
3 years ago
13

A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotat

es horizontally about one end while the pilot is seated at the other end. In order to achieve a radial acceleration of 27.9 m/s2 with a beam of length 5.21 m , what rotation frequency is required
Physics
1 answer:
slava [35]3 years ago
8 0

Answer: the angular frequency is 2.31 rad/s

Explanation:

The data we have is:

Radial acceleration A = 27.9 m/s^2

Beam length r = 5.21m

The radial acceleration is equal to the velocity square divided the radius of the circle (the lenght of the beam in this case)

And we can write the velocity as:

v = w*r where r is the radius of the circle, and w is the angular frequency.

w = 2pi*f

where f is the "normal" frequency.

So we have:

A = (v^2)/r = (r*w)^2/r = r*w^2

We can replace the values and find w.

27.9m/s^2 = 5.21m*w^2

√(27.9/5.21) = w = 2.31 rad/s

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Learning Goal:
Rudiy27

Answer:

The questions are not complete so this is the complete questions

1. How much work W does the motor do on the platform during this process?

2. What is the angular velocity ωf of the platform at the end of this process?

3. What is the rotational kinetic energy, Ek, of the platform at the end of the process described above?

4. How long does it take for the motor to do the work done on the platform calculated in Part 1?

5. What is the average power delivered by the motor in the situation above?

6. . Note that the instantaneous power P delivered by the motor is directly proportional to ω, so P increases as the platform spins faster and faster. How does the instantaneous power P•f being delivered by the motor at the time t•f compare to the average power

P(average) calculated in Part e?

Explanation:

Given that,

The torque τ=25Nm

Moment of inertia I =50kgm²

The platform is initially at rest,

ω•i=0 rad/sec

Revolution the torque produce is 12

Then, θ=12 revolution

1 revolution=2πrad

So, θ=24πrad

1. Work done in a rotational motion is give as

W=τ•Δθ

Given that the τ=25Nm and the initial angular displacement is 0rad

The final angular displacement is 24πrad

Δθ =(θ2-θ1)

Δθ=24π-0

Δθ=24πrad

Then,

W=τ•Δθ

W=25(24π)

W=25×24π

W=1884.96J

To 4s.f, W=1885J

2. Final angular velocity ωf

Using the angular equation

ω•f²=ω•i²+2•α•Δθ

We need to get angular acceleration

The torque is given as

τ=I•α

Given that,

I is moment if inertia =50kgm²

τ=25Nm

α=τ/I

α= 25/50

α=0.5rad/s²

Now, using the angular acceleration

ω•f²=ω•i²+2•α•Δθ

ω•f²=0²+2×0.5×24π

ω•f²=0+75.398

ω•f²=75.398

ω•f=√75.398

ω•f=8.68 rad/sec.

3. We need to find rotational Kinetic energy and it is given as

K.E, = ½I•ω²

Given that, I=50kgm² and ω•f=8.68rad/sec

Then,

K.E, =½I•ω²

K.E, =½×50×8.68²

K.E, =1884.96J

To 4s.f,

K.E, =1885J

Which is the same as the work done by the motor.

4. Time taken to complete part 1,

Using the rotational equation

ω•f=ω•i+α•t

Since, ω•f=8.68 rad/sec and ω•i=0

And α=0.5rad/s²

Then,

ω•f=ω•i+α•t

8.68=0+0.5t

8.68=0.5t

Then, t=8.68/0.5

t=17.36secs

5. The average power of rotational motion is given as

P(average) =Workdone/timetaken

Since,

Work done =1884.96J

Time taken =17.36sec

P(average) =Workdone/timetaken

P(average)=1884.96/17.36

P(average)= 108.58Watts

To 4s.f

P(average)=108.6Watts

6. We need to find •, it is given as

• =τ•ωf

Given that, ω•f=8.68rad/sec, τ=25Nm

•=25×8.68

•=217Watts

Then, the ratio of • to P(average) is

Ratio = •/ P(average)

Ratio= 217/108.58

Ratio=1.9985

Then, the ratio is approximately 2

Ratio=2

5 0
4 years ago
1 ) when a ball is projected upwords its time of rising is ...............the time of falling .
balu736 [363]

Answer:

correct answer is C

Explanation:

In this exercise, you are asked to complete the sentences so that the sentence makes sense.

1) in projectile launching, the only force that acts is gravity in the vertical direction, so the time of going up is EQUAL to the time of going down

correct answer C

2) when a body falls freely, the acceleration is the ratio of gravity, therefore if it starts from rest, its height is

            y = v₀ t - ½ gt²

v₀ = 0

             y = -1/2 g t²

so the position is not proportional to the square of the time

correct answer is C

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3 years ago
Investigate the relationship between volume and pressure of a gas at a constant temperature.
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Gay Lussac's Law states: At a constant volume Pressure<span> divided by </span>Temperature<span> is</span>constant<span> P/T = k Together these three laws form the foundation of the Ideal </span>Gas<span>Law. Objective: Students will </span>investigate<span> Gay Lussac's Law relating </span>pressure<span> and</span>temperature<span> at a </span><span>constant temperature.</span>
7 0
3 years ago
In a race, a runner traveled 12 meters in 4.0 seconds as she accelerated uniformly from rest. The magnitude of the acceleration
mestny [16]
<span>The runner is moving by uniformly accelerated motion, starting from rest (so, his initial velocity is zero). The law of motion of the runner is
</span>x(t) =  \frac{1}{2}  at^2<span>
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</span>a= \frac{2S}{t^2}<span>
We know the runner has covered a distance of S=12m in t=4.0 s, and if we plug these numbers into the equation, we find the acceleration of the runner: 
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3 years ago
If the temperature of the metal oxide and water solution is increased,this would most likely..
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Chemical Reaction between metal oxide and water solution
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