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kherson [118]
3 years ago
6

What is simple machines and it type.I need it today please​

Physics
1 answer:
raketka [301]3 years ago
4 0

Simple machines are tools, devices or objects that are used to make work easier.

TYPES OF SIMPLE MACHINES.

  • Levers
  • Inclined Plane
  • Wedges
  • Screw
  • Pulley
  • Wheel & Axle

<u>T</u><u>H</u><u>E</u><u> </u><u>L</u><u>E</u><u>V</u><u>E</u><u>R</u><u>S</u>

A lever is simply a plank or ridge beam that is free to rotate on a pivot.

<u>P</u><u>A</u><u>R</u><u>T</u><u>S</u><u> </u><u>O</u><u>F</u><u> </u><u>A</u><u> </u><u>L</u><u>E</u><u>V</u><u>E</u><u>R</u><u> </u>

  1. Load
  2. Effort
  3. Fulcrum or pivot

<u>C</u><u>L</u><u>A</u><u>S</u><u>S</u><u>E</u><u>S</u><u> </u><u>O</u><u>F</u><u> </u><u>L</u><u>E</u><u>V</u><u>E</u><u>R</u><u>S</u>

THE FIRST CLASS LEVER

In this class, the FULCRUM is between the EFFORT and the LOAD. The mechanical advantage is more if the load is closer to the FULCRUM. Examples: seesaws, boat oars and crowbars.

THE SECOND CLASS LEVER

In this class, the LOAD is between the EFFORT and the FULCRUM. The mechanical advantage is more if the LOAD is closer to the FULCRUM. Example: wheelbarrow.

THE THIRD CLASS LEVER

In this class, the EFFORT is between the LOAD and the FULCRUM. The mechanical advantage is more if the EFFORT is closer to the LOAD. Example: garden shovel.

<u>T</u><u>H</u><u>E</u><u> </u><u>INCLINED</u><u> </u><u>PLANE</u><u> </u><u>OR</u><u> </u><u>RAMP</u>

An inclined plane is a simple machine with a sloping surface. It makes it easier for us to move objects from lower grounds to higher grounds.

<u>W</u><u>E</u><u>D</u><u>G</u><u>E</u><u>S</u>

A wedge is simply a triangular tool, often made of wood, stone, metal or plastic. It is thick at one end and tapers to a thin or sharp edge at the other end. Examples: knife, axe, doorstopper, nail, blade on the snow plough or farmer grader.

<u>PULLEY</u>

A pulley is simply a wheel with a groove in it, and a rope in the groove. It is used to lift heavy objects from the ground to high places.

<u>WHEEL</u><u> </u><u>&</u><u> </u><u>AXLE</u>

Involves two circular objects joined at the center. It works when force is applied to the wheel or when force is applied to the axle. Example: door knob.

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puteri [66]
Like that the heat of the room doesn't reach the what's inside the sealed container. Also for the sealed container to not let out the heat inside the container
7 0
4 years ago
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A truck exerts 2700 N of force on a car that is stuck in the mud, and does not move. How much work has the truck performed?
vova2212 [387]

Answer:

zero

Explanation:

although force was applied as long as there's no movement ,no work has been done

mathematically;work=force×distance

work=2700×0

work=0J

hope I helped

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5 0
4 years ago
A force of 500N Is applied to a steel wire of goss_sectional area 0.2m2 the tensele stress is a. 2.5x10 1nm-2 b. 1 0x102nm-2 c.
Arturiano [62]

Answer:

the tensile stress of the steel wire is 2.5 x 10³ Nm⁻².

Explanation:

Given;

applied force, F = 5000 N

cross sectional area of the steel wire, A = 0.2 m²

The tensile stress of the steel wire, is calculated as;

\sigma = \frac{F}{A} \\\\\sigma = \frac{500}{0.2} \\\\\sigma = 2500 \ N/m^2\\\\\sigma = 2.5 \ \times \ 10^3 \ Nm^{-2}

Therefore, the tensile stress of the steel wire is 2.5 x 10³ Nm⁻².

3 0
3 years ago
1. What is the pull that all objects exert on each other?
Andrei [34K]

Answer:

Im pretty sure 1 is gravity, 2 is force

Explanation:

8 0
4 years ago
Completenlo por favor
Ainat [17]

La velocidad \mathbf v del objeto al tiempo t es descrito por

\mathbf v(t)=v_x(t)\,\mathbf i+v_y(t)\,\mathbf j

donde

\begin{cases}v_x(t)={v_x}_0+a_xt\\v_y(t)={v_y}_0+a_yt\end{cases}

El objeto no tiene aceleración horizontal, pues a_x=0 y v_x está determinado exactamente por su velocidad inicial en la dirección horizontal. En la dirección vertical, la gravedad es la única fuerza que actúa en el objeto, pues a_y=-9.81\,\dfrac{\mathrm m}{\mathrm s^2}. Entonces, la velocidad después de 3\,\mathrm s satisface

{v_x}_0\,\mathbf i+\left({v_y}_0+\left(-9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)(3\,\mathrm s)\right)\,\mathbf j=20\,\mathbf i-4\,\mathbf j

Inmediatamente, vemos que {v_x}_0=20\,\dfrac{\mathrm m}{\mathrm s} y podemos encontrar que {v_y}_0=25.43\,\dfrac{\mathrm m}{\mathrm s}.

Su posicíon es descrita al tiempo t por

\mathbr r(t)=r_x\,\mathbf i+r_y\,\mathbf j

con

\begin{cases}r_x={r_0}_x+{v_0}_xt+\dfrac12a_xt^2\\\\r_y={r_0}_y+{v_0}_yt+\dfrac12a_yt^2\end{cases}

Si la posición inicial del objeto es el origen, suponemos que \mathbr r(0)=\mathbf 0, y además tenemos

\mathbf r(t)=\left(20\,\dfrac{\mathrm m}{\mathrm s}\right)t\,\mathbf i+\left(\left(25.43\,\dfrac{\mathrm m}{\mathrm s}\right)t+\dfrac12\left(-9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\right)\,\mathbf j

Queremos determinar el máximo de r_y. Encontramos que {r_y}_{\mathrm{max}}\approx32.9\,\mathrm m cuando t\approx2.59\,\mathrm s.

3 0
3 years ago
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