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tino4ka555 [31]
3 years ago
10

He __________ uses a series of gears to adjust the output speed of the engine.

Physics
1 answer:
Verizon [17]3 years ago
4 0
Hey /人 ◕ ‿‿ ◕ 人\
The answer is transmission

uses a series of gears to transmit power to facilitate changes in speed .


GLAD TO HELP


~~~ ╔͎═͓═͙╗
~~~ ╚̨̈́═̈́﴾ ̥̂˖̫˖̥  ̂ )
You might be interested in
A surface receiving sound is moved from its original position to a position three times farther away from the source of the soun
vazorg [7]

Answer:

c. nine times as low.

Explanation:

Sound intensity is defined as the acoustic power transferred by a sound wave per unit of normal area to the direction of propagation:

I\propto \frac{1}{A}

Since the sound wave has a spherical wavefront of radius r, then the area is given by:

A=4\pi r^2

Here r is the distance from the source of the sound. Thus sound intensity decreases as:

I\propto \frac{1}{4\pi r^2}\\\\I'\propto \frac{1}{4\pi r'^2}\\\\I'\propto \frac{1}{4\pi (3r)^2}\\\\I'\propto \frac{1}{9} \frac{1}{4\pi r^2}\\\\I'\propto \frac{1}{9} I

6 0
3 years ago
A projectile of mass 2.0 kg is fired in the air at an angle of 40.0 ° to the horizon at a speed of 50.0 m/s. At the highest poin
tekilochka [14]

Answer:

a) The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

                                         0.7 kg is 109.4 ms on the x axis

b)  Y = 109.3 m

Explanation:

This is a moment and projectile launch exercise.

a) Let's start by finding the initial velocity of the projectile

       sin 40 = voy / v₀

       v_{oy} = v₀ sin 40

       v_{oy} = 50.0 sin40

       v_{oy} = 32.14 m / s

       cos 40 = v₀ₓ / V₀

       v₀ₓ = v₀ cos 40

       v₀ₓ = 50.0 cos 40

       v₀ₓ = 38.3 m / s

Let us define the system as the projectile formed t all fragments, for this system the moment is conserved in each axis

Let's write the amounts

Initial mass of the projectile M = 2.0 kg

Fragment mass 1 m₁ = 1.0 kg and its velocity is vₓ = 0 and v_{y} = -10.0 m / s

Fragment mass 2 m₂ = 0.7 kg moves in the x direction

Fragment mass 3 m₃ = 0.3 kg moves up (y axis)

Moment before the break

X axis

     p₀ₓ = m v₀ₓ

Y Axis y

    p_{oy} = 0

After the break

X axis

   p_{fx} = m₂ v₂

Axis y

     p_{fy} = m₁ v₁ + m₃ v₃

Let's write the conservation of the moment and calculate

Y Axis  

     0 = m₁ v₁ + m₃ v₃

Let's clear the speed of fragment 3

     v₃ = - m₁ v₁ / m₃

     v₃ = - (-10) 1 / 0.3

     v₃ = 33.3 m / s

X axis

     M v₀ₓ = m₂ v₂

     v₂ = v₀ₓ M / m₂

     v₂ = 38.3  2 / 0.7

     v₂ = 109.4 m / s

The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

                                         0.7 kg is 109.4 ms on the x axis

b) The speed of the fragment is 33.3 m / s and has a starting height of where the fragmentation occurred, let's calculate with kinematics

       v_{fy}² = v_{oy}² - 2 gy

       0 =  v_{oy}²-2gy

       y =  v_{oy}² / 2g

       y = 32.14² / 2 9.8

       y = 52.7 m

This is the height where the break occurs, which is the initial height for body movement of 0.3 kg

      v_{f}² =  v_{y}² - 2 g y₂

      0 =  v_{y}² - 2 g y₂

     y₂ =  v_{y}² / 2g

     y₂ = 33.3²/2 9.8

     y₂ = 56.58 m

Total body height is

      Y = y + y₂

      Y = 52.7 + 56.58

     Y = 109.3 m

8 0
2 years ago
Which has the greater acceleration, a person going from 0 m/s to 10 m/s in 10 seconds or an ant going from 0 m/s to 0.25 m/s in
monitta

<u>P</u><u>e</u><u>r</u><u>s</u><u>o</u><u>n</u><u>-</u><u>1</u>

  • Initial velocity=u=0m/s
  • Final velocity=v=10m/s
  • Time=10s=t

\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}

\\ \sf\longmapsto Acceleration=\dfrac{10-0}{10}

\\ \sf\longmapsto Acceleration=\dfrac{10}{10}

\\ \sf\longmapsto Acceleration=1m/s^2

<u>P</u><u>e</u><u>r</u><u>s</u><u>o</u><u>n</u><u>-</u><u>2</u>

  • initial velocity=0m/s=u
  • Final velocity=v=0.25m/s
  • Time=t=2s

\\ \sf\longmapsto Acceleration=\dfrac{0.25-0}{2}

\\ \sf\longmapsto Acceleration=\dfrac{0.25}{2}

\\ \sf\longmapsto Acceleration=0.125m/s^2

Person-1 is accelerating faster.

4 0
3 years ago
What makes these image a represent of what research <br>​
ycow [4]
The economic down fall
3 0
3 years ago
Is it proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinde
Gelneren [198K]

Answer:

No, it is not proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinder.

Explanation:

A cylinder is said to be infinitely long when is of a sufficient length. Also, when the diameter of the cylinder is relatively small compared to the length, it is called infinitely long cylinder.

Cylindrical rods can also be treated as infinitely long when dealing with heat transfers at locations far from the top or bottom surfaces. However, it not proper to treat the cylinder as being infinitely long when:

* When the diameter and length are comparable (i.e have the same measurement)

When finding the temperatures near the bottom or top of a cylinder, it is NOT PROPER TO USE AN INFINITELY LONG CYLINDER because heat transfer at those locations can be two-dimensional.

Therefore, the answer to the question is NO, since it is not proper to use an infinitely long cylinder when finding temperatures near the bottom or top of a cylinder.

8 0
3 years ago
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