Agree or disagree

*photo attached* The diameters of the main rotor and tail rotor of a single-engine helicopter are 7.67 m and 1.01 m, respectivel

stiks02 [169]

(a) The speeds of the tips of both rotors; main rotor <u>178.3 m/s</u> and tail rotor <u>218.4 m/s</u>.

(b) The speed of the main rotor is <u>0.52</u> speed of sound, and the speed of the tail rotor is <u>0.64</u> speed of sound.

<h3>Linear speed of main motor and tail rotor</h3>

v = ωr

where;

ω is the angular speed (rad/s)
r is radius (m)

v(main rotor) = (444 rev/min x 2π rad x 1 min/60s) x (0.5 x 7.67 m)

v(main rotor) = 178.3 m/s

v(tail rotor) = (4,130 rev/min x 2π rad x 1 min/60s) x (0.5 x 1.01 m)

v(tail rotor) = 218.4 m/s

<h3>Speed of the rotors with respect to speed of sound</h3>

% speed (main motor) = 178.3/343 = 0.52 = 52 %

% speed (tail motor) = 218.4/343 = 0.64 = 64 %

Thus, the speed of the main rotor is 0.52 speed of sound, and the speed of the tail rotor is 0.64 speed of sound.

Learn more about linear speed here: brainly.com/question/15154527

#SPJ1

5 0

2 years ago

A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this

wlad13 [49]

Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

$r=r_o\times A^{\frac{1}{3}}$

$r_o=1.25 \times 10^{-15} m$ = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

$r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}$

$r=4.6656\times 10^{-15} m=4.6656 fm$

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) $F=k\times \frac{q_1q_2}{a^2}$

k= $9\times 10^9 N m^2/C^2$ = Coulombs constant

$q_1,q_2$ = charges kept at distance 'a' from each other

F = electrostatic force between charges

$q_1=+1.602\times 10^{-19} C$

$q_2=+1.602\times 10^{-19} C$

Force of repulsion between two protons on opposite sides of the diameter

$a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m$

$F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}$

$F=2.6527 N$

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

6 0

3 years ago

Select the correct answer.

Ilia_Sergeevich [38]

Pretty sure it is B.
Because inertia is a tendency to do nothing or remain unchanged.

8 0

3 years ago

Vector A has magnitude 8.00 mm and is in the xy-plane at an angle of 127∘ counterclockwise from the +x–axis (37∘ past the +y-axi

Ilia_Sergeevich [38]

Answer:

-75.35°

Explanation:

Let C be the sum of the two vectors A and B. Hence, we can write the following

$A_{x} +B_{x} =C_{x} ......(1)\\A_{y} +B_{y} =C_{y} ......(2)$

but since the vector C is in the -y direction, $C_{x}$ = 0 and $C_{y}$ = —12 m.

Thus

$B_{x} =-A_{x} =-[-Acos(180-127)]=(8)*cos(53)\\B_{x} =4.81m$

similarly, we can determine $B_{y}$ by rearranging equation (1)

$B_{y} =C_{y} -A_{y} =-12m-[(8)*sin(53)\\B_{y} =-18.4m$

so the magnitude of B is

$B=\sqrt{B_{x}^2+B_{y}^2 } \\B=19m$

Finally, the direction of B can be calculated as follows

Ф= $tan^{-1} (\frac{B_{y} }{B_{x} } )\\=-75.35$

hence the vector B makes an angle of 75.35 clockwise with + x axis

8 0

4 years ago

Q4. Consider the skier on a slope shown in the figure below. Her mass including equipment is 55.0 kg.

Shkiper50 [21]

Answer:

Part a)

When there is no friction then acceleration is

$a = 4.14 m/s^2$

Part b)

if there is friction force along the inclined then acceleration is

$a = 3.33 m/s^2$

Explanation:

Part a)

As we know that the skier is on inclined plane

So here if there is no friction then net force along the inclined plane is given as

$F = mg sin\theta$

now acceleration of the skier is given as

$a = \frac{F}{m}$

$a = g sin\theta$

$a = 9.81(sin25)$

$a = 4.14 m/s^2$

Part b)

if there is friction force along the inclined then net force along the inclined plane is given as

$F = mg sin\theta - F_f$

now acceleration of the skier is given as

$a = \frac{F}{m}$

$a = g sin\theta - \frac{F_f}{m}$

$a = 9.81(sin25) - \frac{45}{55}$

$a = 3.33 m/s^2$

5 0

4 years ago