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ANEK [815]
3 years ago
12

The weight of the North American P-51 Mustang airplane is 10,100 lb and its wing platform area is 233 ft2 . Calculate the wing l

oading (airplane weight per wing area) in both English engineering and SI units. Also, express the wing loading in terms of the nonconsistent (but commonly used) unit kgf/m2 (kilogram-force per square meter).
Physics
1 answer:
Natali5045456 [20]3 years ago
5 0

Answer

given,

weight of airplane,W = 10,100 lb

Wing area,A = 233 ft²

now,

Wing Loading Calculation in English engineering

        = \dfrac{W}{A}

        = \dfrac{10,100}{233}

        = 43.35 lb/ft²

Wing Loading in SI unit

        = 43.35\times \dfrac{4.448\ N}{1\ lb}\times (\dfrac{1\ ft}{0.3048})^2

        = 2075.51\ N/m^2

Wing loading in kgf/m²

      = \dfrac{2075.51}{9.81}

      = 211.57 kgf/m²

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Brut [27]

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The car's momentum was (mass) x (speed)

Momentum = (2400 kg) x (20 m/s)

Momentum = 48,000 km-m/s

To completely stop the car, the impulse = -48,000 km-m/s .

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3 years ago
Two tuning forks, 254 Hz. and 260 Hz., are struck simultaneously. How many beats will be heard?
Lisa [10]

The longer you continue to listen, the more beats will be heard.

They'll occur at the rate of (260Hz - 254Hz) = 6 Hz .
 
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3 years ago
A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly
Mama L [17]

Answer:

Explanation:

To solve this, we start by using one of the equations of motion. The very first one, in fact

1

V = U + at.

V = 0 + 0.8 * 3.4 = 2.72 m/s.

2.

V = 0 + 0.8 * 4.3 = 3.44 m/s.

3.

d = ½ * 0.8 * 4.3² + 3.44 * 12.9

d = 7.396 + 44.376

d = 51.77 m.

4.

d = 62 - 51.77 = 10.23 m. = Distance

traveled during deceleration.

a = (V² - Vo²) / 2d.

a = (0² - 3.44²) / 20.46

a = -11.8336 / 20.46 = -0.58 m/s²

5.

t = (V - Vo)/a =(0 - 3.44) / -0.58

t = -3.44/-.58 = 5.93 s

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T = 4.3 + 12.9 + 5.93 = 23.13 s. = Total

time the hare was moving.

6.

d = Vo * t + ½ * a * t² = 62 m.

0 + 0.5 * (23.13)² * a = 61

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7 0
3 years ago
a ballistic pendulum is used to measure the speed of high-speed projectiles. A 6 g bullet A is fired into a 1 kg wood block B su
Galina-37 [17]

Answer:

(a) v-bullet = 399.04 m/s

(b) I = 2.38 kg m/s

(c) T = 2.59 N

Explanation:

(a) To calculate the initial speed of the bullet, you first take into account that the kinetic energy of both wood block and bullet, just after the bullet impacts the block, is equal to the potential gravitational energy of block and bullet when the cord is at 60° respect to the vertical.

The potential energy is given by:

U=(M+m)gh       (1)

U: potential energy

M: mass of the wood block = 1 kg

m: mass of the bullet = 6g = 6.0*10^-3 kg

g: gravitational constant = 9.8m/s^2

h: distance to the ground

The distance to the ground is calculate d by using the information about the length of the cord and the degrees of the cord respect to the vertical:

h=l-lsin\theta\\\\h=2.2m-2,2m\ sin60\°=0.29m

The potential energy is:

U=(1kg+6*10^{-3}kg)(9.8m/s^2)(0.29m)=2.85J

Next, the potential energy is equal to kinetic energy of the block and the bullet at the beginning of its motion:

U=\frac{1}{2}(M+m)v^2\\\\v=\sqrt{2\frac{U}{M+m}}=\sqrt{2\frac{2.85J}{1kg+6*10^{-3}kg}}=2.38\frac{m}{s}

Next, you use the momentum conservation law, in order to calculate the speed of the bullet before the impact:

Mv_1+mv_2=(M+m)v    (2)

v1: initial velocity of the wood block = 0m/s

v2: initial speed of the bullet

v: speed of bullet and block = 2.38m/s

You solve the equation (2) for v2:

M(0)+mv_2=(M+m)v    

v_2=\frac{M+m}{m}v=\frac{1kg+6*10^{-3}kg}{6*10^{-3}kg}(2.38m/s)\\\\v_2=399.04\frac{m}{s}

The speed of the bullet before the impact with the wood block is 399.04 m/s

(b) The impulse is gibe by the change in the velocity of the block, multiplied by the mass of the block:

I=M\Delta v=M(v-v_1)=(1kg)(2.38m/s-0m/s)=2.38kg\frac{m}{s}

The impulse is 2.38 kgm/s

(c) The force on the cord after the impact is equal to the centripetal force over the block and bullet. That is:

T=F_c=(M+m)\frac{v^2}{l}=(1.006kg)\frac{(2.38m/s)^2}{2.2m}=2.59N    

The force on the cord after the impact is 2.59N

4 0
3 years ago
Look at the equation. What detail is missing? 3 m/s2= (33 m/s - X)/30 S <br>​
Tems11 [23]

Answer:

The starting velocity.

Explanation:

We must understand that this equation comes from the following equation of kinematics.

v_{f}=v_{o}+a*t

where:

Vf = final velocity = 33 [m/s]

Vo = starting velocity [m/s]

a = acceleration = 3 [m/s²]

t = time = 30 [s]

So, these values can be assembly in the following way:

v_{f}=v_{o}+a*t\\a*t=v_{f}-v_{o}\\3=\frac{33-v_{o}}{30}

6 0
3 years ago
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