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VladimirAG [237]

3 years ago

15

A typical small rescue helicopter has four blades, each is 4.00 m long and has a mass of 50.0 kg. The blades can be approximated

as thin rods that rotate about one end of an axis perpendicular to their length. The helicopter has a total loaded mass of 1000 kg (including the four blades). (a) Calculate the rotational kinetic energy in the blades as they rotate at 300 rev/min.

1 answer:

lisov135 [29]3 years ago

8 0

Answer:

I1 = 1/3 m L^2 = 50 * 4^2 / 3 = 267 kg-m^2 inertia of one blade

I = 4 * I1 = 1067 kg-m^2 inertia of rotor

f = 300/min = 5 / sec

ω = 2 pi * f = 31.4 / sec

KE = 1/2 * I * ω^2 = 526,000 J energy of rotor

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La frecuencia de una onda es 60Hz y su velocidad, 30 m/s. Calcula su longitud de onda

Andreas93 [3]

   Wavelength = (speed) / (frequency)

   = (30 m/s) / (60/sec) =

   = 0.5 meter .

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4 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

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6 0

3 years ago

A large, 68.0-kg cubical block of wood with uniform density is floating in a freshwater lake with 20.0% of its volume above the

LenaWriter [7]

Answer:

a) V = 0.085 m^3

b) m = 17 kg

Explanation:

1) Data given

mb = 68 kg (mass for the block)

20% of the block volume is floating

100-20= 80% of the block volume is submerged

2) Notation

mb= mass of the block

Vw= volume submerged

mw = mass water displaced

V= total volume for the block

3) Forces involved (part a)

For this case we have two forces the buoyant force (B), defined as the weight of water displaced acting upward and the weight acting downward (W)

Since we have an equilibrium system we can set the forces equal. By definition the buoyant force is given by :

B = (mass water displaced) g = (mw) g (1)

The definition of density is :

$\rho_w = \frac{m_w}{V_w}$

If we solve for mw we got $m_w = \rho_w V_w$ (2)

Replacing equation (2) into equation (1) we got:

$B = \rho_w V_w g$ (3)

On this case Vw represent the volume of water displaced = 0.8 V

If we replace the values into equation (3) we have

0.8 ρ_w V g = mg (4)

And solving for V we have

V = (mg)/(0.8 ρ_w g )

We cancel the g in the numerator and the denominator we got

V = (m)/(0.8 ρ_w)

V = 68kg /(0.8 x 1000 kg/m^3) = 0.085 m^3

4) Forces involved (part b)

For this case we have bricks above the block, and we want the maximum mass for the bricks without causing it to sink below the water surface.

We can begin finding the weight of the water displaced when the block is just about to sink (W1)

W1 = ρ_w V g

W1 = 1000 kg/m^3 x 0.085 m^3 x 9.8 m/s^2 = 833 N

After this we can calculate the weight of water displaced before putting the bricks above (W2)

W2 = 0.8 x 833 N = 666.4 N

So the difference between W1 and W2 would represent the weight that can be added with the bricks (W3)

W3 = W1 -W2 = 833-666.4 N = 166.6 N

And finding the mass fro the definition of weight we have

m3 = (166.6 N)/(9.8 m/s^2) = 17 Kg

8 0

4 years ago

Jupiter has a mass of 1,898,000,000,000,000,000,000,000,000 kg. How would

Natasha2012 [34]

Answer:

B.1.898^27 kg

Explanation:

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4 years ago

A crane raises a crate with a mass of 150 kg to a height of 20 m. Given that

Virty [35]

Answer:

$\boxed {\boxed {\sf 29,400 \ Joules}}$

Explanation:

Gravitational potential energy is the energy an object possesses due to its position. It is the product of mass, height, and acceleration due to gravity.

$E_P= m \times g \times h$

The object has a mass of 150 kilograms and is raised to a height of 20 meters. Since this is on Earth, the acceleration due to gravity is 9.8 meters per square second.

m= 150 kg
g= 9.8 m/s²
h= 20 m

Substitute the values into the formula.

$E_p= 150 \ kg \times 9.8 \ m/s^2 \times 20 \ m$

Multiply the three numbers and their units together.

$E_p=1470 \ kg*m/s^2 \times 20 m$

$E_p=29400 \ kg*m^2/s^2$

Convert the units.

1 kilogram meter square per second squared (1 kg *m²/s²) is equal to 1 Joule (J). Our answer of 29,400 kg*m²/s² is equal to 29,400 Joules.

$E_p= 29,400 \ J$

The crate has <u>29,400 Joules</u> of potential energy.

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3 years ago