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aniked [119]
3 years ago
10

If each of the three rotor helicopter blades is 3.50 m long and has a mass of 120 kg , calculate the moment of inertia of the th

ree rotor blades about the axis of rotation.
Physics
1 answer:
devlian [24]3 years ago
7 0

Answer:

1470kgm²

Explanation:

The formula for expressing the moment of inertial is expressed as;

I = 1/3mr²

m is the mass of the body

r is the radius

Since there are three rotor blades, the moment of inertia will be;

I = 3(1/3mr²)

I = mr²

Given

m = 120kg

r = 3.50m

Required

Moment of inertia

Substitute the given values and get I

I = 120(3.50)²

I = 120(12.25)

I = 1470kgm²

Hence the moment of inertial of the three rotor blades about the axis of rotation is 1470kgm²

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If a person is walking along at 1.4m/s. How long will it take that person to walk one time around a high school track?​
ratelena [41]

Answer:

285.7s

Explanation:

Given parameters:

Speed of the person  = 1.4m/s

Unknown:

Time it takes to walk one time round the track = ?

Solution:

The circumference of the track is 400m for a standard pitch.

Now;

 Speed  = \frac{distance}{time}

 Time  = \frac{distance }{speed }  

 Now insert the parameters;

Time  = \frac{400}{1.4}  = 285.7s

8 0
3 years ago
In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.46 m. The mug
Tresset [83]

Answer:

(a). The horizontal velocity is 1.46 m/s.

(b). The direction of the mug's velocity just before it hit the floor is 74.7° below the horizontal.

Explanation:

Given that,

Height of the counter = 1.46 m

Distance = 0.80 m

We need to calculate the time

Using equation of motion

s_{y}=ut+\dfrac{1}{2}gt^2

s_{y}=\dfrac{1}{2}gt^2

Put the value into the formula

1.46=\dfrac{1}{2}\times9.8\times t^2

t^2=\dfrac{1.46\times 2}{9.8}

t=\sqrt{\dfrac{1.46\times2}{9.8}}

t=0.545\ sec

Here, horizontal velocity is constant

(a). We need to calculate the velocity

Using formula of velocity

v_{x}=\dfrac{d}{t}

Put the value into the formula

v_{x}=\dfrac{0.80}{0.545}

v_{x}=1.46\ m/s

(b). We need to calculate the final velocity

Using equation of motion

v_{f}^2=u^2+2as

Put the value into the formula

v_{f}^2=0+2\times9.8\times1.46

v_{f}=\sqrt{28.616}

v_{f}=5.34\ m/s

The velocity is 5.34 m/s downward.

We need to calculate the direction

Using formula of direction

\tan\theta=\dfrac{v_{y}}{v_{x}}

\theta=\tan^{-1}(\dfrac{5.34}{1.46})

\theta=74.7^{\circ}

Hence, (a). The horizontal velocity is 1.46 m/s.

(b). The direction of the mug's velocity just before it hit the floor is 74.7° below the horizontal.

5 0
4 years ago
A baseball is hit almost straight up into the air with a speed of 26 m/s . Estimate how high it goes.
Natalka [10]

Answer:

The maximum height of the ball is 34.5 m.

The ball is 5.31 s in the air.

Explanation:

Hi there!

The equations for the height and velocity of the baseball that is hit straight up are as follows:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the baseball at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g =  acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity at time t.

If we place the origin of the frame of reference at the place where the baseball is hit, then, y0 = 0.

To calculate how high it goes, we have to obtain the time at which the ball is at maximum height. At that point, the velocity is 0. Then using the equation of velocity:

v = v0 + g · t

0 = 26 m/s - 9.8 m/s² · t

-26 m/s / -9.8 m/s² = t

t = 2.65 s

The height at that time will be the maximum height:

y = y0 + v0 · t + 1/2 · g · t²        (y0 = 0)

y = 26 m/s · 2.65 s - 1/2 · 9.8 m/s² · (2.65 s)²

y = 34.5 m

The maximum height of the ball is 34.5 m

If it takes the ball 2.65 s to reach the maximum height it will take another 2.65 s to return to the initial position. Then, the time it will be in the air is (2.65 s + 2.65 s) 5.30 s. However, let´s calculate the time it takes the ball to reach the initial position using the equation for height.

At the initial position y = 0. Then:

y = y0 + v0 · t + 1/2 · g · t²        (y0 = 0)

0 = 26 m/s · t - 1/2 · 9.8 m/s² · t²

0 = t (26 m/s - 1/2 · 9.8 m/s² · t)      (t = 0 when the ball is hit)

0 = 26 m/s - 1/2 · 9.8 m/s² · t

-26 / -4.9 m/s² = t

t = 5.31 s     ( the difference with the 5.30 s obtained above is due to rounding the time to 2.65 s).

The ball is 5.31 s in the air.

Have a nice day!

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3 years ago
Which kinds of objects emit light? A. objects that shine B. hot objects C. distant objects D. colored objects E. all objects
Vinil7 [7]

Answer:

Correct answer is A. object that shine

3 0
4 years ago
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pogonyaev
<em>FORCE IS EXPRESSED IN NEWTONS.</em>
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