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Zanzabum
3 years ago
5

The major horizontal surface on an airplane that provides lift.​

Physics
1 answer:
fgiga [73]3 years ago
5 0

wings

Lift is generated by every part of the airplane, but most of the lift on a normal airliner is generated by the wings. Lift is a mechanical aerodynamic force produced by the motion of the airplane through the air.

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When sweat evaporates from your skin the temp of your skin decreases this is because
Vanyuwa [196]
The water molecules with a slower speed are escaping
8 0
3 years ago
A mass m attached to a horizontal massless spring with spring constant k, is set into simple harmonic motion. its maximum displa
Lesechka [4]
At the point of maximum displacement (a), the elastic potential energy of the spring is maximum:
U_i= \frac{1}{2} ka^2
while the kinetic energy is zero, because at the maximum displacement the mass is stationary, so its velocity is zero:
K_i =0
And the total energy of the system is
E_i = U_i+K= \frac{1}{2}ka^2

Viceversa, when the mass reaches the equilibrium position, the elastic potential energy is zero because the displacement x is zero:
U_f = 0
while the mass is moving at speed v, and therefore the kinetic energy is
K_f =  \frac{1}{2} mv^2
And the total energy is
E_f = U_f + K_f =  \frac{1}{2} mv^2

For the law of conservation of energy, the total energy must be conserved, therefore E_i = E_f. So we  can write
\frac{1}{2} ka^2 =  \frac{1}{2}mv^2
that we can solve to find an expression for v:
v= \sqrt{ \frac{ka^2}{m} }
6 0
3 years ago
A shovel is the third class lever​
Nookie1986 [14]

Answer:

Yes

Explanation:

In a third-class lever, the effort force lies between the resistance force and the fulcrum. Some kinds of garden tools are examples of third-class levers. When you use a shovel, for example, you hold one end steady to act as the fulcrum, and you use your other hand to pull up on a load of dirt.

8 0
3 years ago
A bald eagle is flying to the left with a speed of 34 meters
Shtirlitz [24]

Answer:

the speed after 3 seconds is 10 m/s

Explanation:

The computation of the speed is shown below:

As we know that

V = U  + at

Here,

U = 34 m/s

a =  - 8 m/s²

t = 3 Sec

V = velocity after 3 sec

V  = 34 + (-8)3

 = 34 - 24

 V = 10 m/s

Hence, the speed after 3 seconds is 10 m/s

4 0
2 years ago
A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for
stealth61 [152]

Answer:

(a) a_s=0.62\frac{m}{s^2}

(b) a_s=0.13\frac{m}{s^2}

(c) x_f=2.6m

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}

(c) Using the kinematics equation:

x_f=x_0+v_0t \pm  \frac{at^2}{2}

For the girl, we have x_0=0 and v_0=0. So:

x_f_g=\frac{a_gt^2}{2}(1)

For the sled, we have v_0=0. So:

x_f_s=x_0_s-\frac{a_st^2}{2}(2)

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s

Now, we solve (1) for x_f_g

x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m

5 0
2 years ago
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