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OleMash [197]
3 years ago
9

Pitch, roll, and yaw are all terms for specific kinds of motion. What does pitch mean?\

Physics
1 answer:
ad-work [718]3 years ago
8 0
All these terms are related to an aircraft system. A pitch is a rotational motion around the side-to-side axis. It is a phenomenon when a plane is tilted upwards to increase lift of the tail or tilted downwards to decrease it. It is controlled by an elevator from the pilot maneuvering system.
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Which two things does force depend on?​
saveliy_v [14]

Answer:

Mass and Gravity

Explanation:

Objects with mass exert forces on each other via the force of gravity. This force is proportional to the mass of the two interacting objects and is inversely proportional to the square of the distance between them.

3 0
3 years ago
What is the speed of sound in air at 40°C? 355 m/s 307 m/s 331 m/s 239 m/s
abruzzese [7]
The ans is A. 355 m/s as temperature is above stp
3 0
3 years ago
Read 2 more answers
Mars has a mass of about 6.24 Ã 1023 kg, and its moon phobos has a mass of about 9.2 Ã 1015 kg. if the magnitude of the gravitat
Effectus [21]
Given: Mass of Mars Mm = 6.24 x 10²³ Kg;  

           Mass of Phobos Mp = 9.2 x 10¹⁵ Kg

           Gravitational  Force F = 4.47 X 10¹⁵ N

           Gravitational constant G = 6.67 X 10⁻¹¹ N m²/Kg²

           Radius r = ?

Formula: F = GMmMp/r²  derive r

               r = √GMmMp/F

     r = √(6.67 x 10⁻¹¹ N m²/Kg²)(6.24 x 10²³ Kg)(9.2 x 10¹⁵ Kg)/4.47 x 10¹⁵ N

     r = √(6.67 x 10⁻¹¹ N m²/Kg²)5.74 x 10³⁹ Kg²/4.47 x 10¹⁵ N

     r = √8.57 x 10¹³ m

     r = 9,257,429 m

or r = 9.26 x 10⁶ m

8 0
4 years ago
43. A rocket sled accelerates at a rate of 49.0m/s2. Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component
lord [1]

(a) 3675 N

Assuming that the acceleration of the rocket is in the horizontal direction, we can use Newton's second law to solve this part:

F_x = m a_x

where

F_x is the horizontal component of the force

m is the mass of the passenger

a_x is the horizontal component of the acceleration

Here we have

m = 75.0 kg

a_x = 49.0 m/s^2

Substituting,

F_x=(75.0)(49.0)=3675 N

(b) 3748 N, 11.3 degrees above horizontal

In this part, we also have to take into account the forces acting along the vertical direction. In fact, the seat exerts a reaction force (R) which is equal in magnitude and opposite in direction to the weight of the passenger:

R=mg=(75.0)(9.8)=735 N

where we used

g=9.8 m/s^2 as acceleration of gravity.

So, this is the vertical component of the force exerted by the seat on the passenger:

F_y = 735 N

and therefore the magnitude of the net force is

F=\sqrt{F_x^2+F_y^2}=\sqrt{3675^2+735^2}=3748 N

And the direction is given by

\theta = tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{735}{3675})=11.3^{\circ}

4 0
3 years ago
Help pls lol and tysm
Zanzabum

Answer:

umm section 2 and 4

Explanation:

because at section 2 it starts and at section 4 it moves again and stops at 3.

7 0
3 years ago
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