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Verdich [7]
3 years ago
15

Draw the diagram of third class lever showing fulcrum, effort and load​

Physics
1 answer:
sesenic [268]3 years ago
8 0

Answer:

In a third class lever, the effort is located between the load and the fulcrum. If the fulcrum is closer to the load, then less effort is needed to move the load. If the fulcrum is closer to the effort, then the load will move a greater distance. ... These levers are useful for making precise movements.

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Indica 3 aplicações dos ultrassons. *
Ainat [17]

Answer:

El ultrasonido se utiliza en muchos campos diferentes. Los dispositivos ultrasónicos se utilizan para detectar objetos y medir distancias. La ecografía o la ecografía se utilizan a menudo en medicina. En las pruebas no destructivas de productos y estructuras, el ultrasonido se utiliza para detectar defectos invisibles.

Espero que esto ayude, tenga un maravilloso día / noche, manténgase a salvo, felices fiestas y feliz navidad.

3 0
3 years ago
Some help me on this please
satela [25.4K]

Answer:

Changes in the object's momentum (answer D)

Explanation:

A net force will cause an object to change its velocity, and that will affect the object's momentum, which is defined by the product of the object's mass times its velocity.

So, select the last option (D) in the given list.

8 0
3 years ago
12. A frain moves from rest to a speed of 25 m/s in 30.0 seconds. What is its acceleration?
ziro4ka [17]
  • initial velocity=u=0m/s
  • Final velocity=v=25m/s
  • Time=t=30s

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

\\ \tt\longmapsto Acceleration=\dfrac{25-0}{30}

\\ \tt\longmapsto Acceleration=\dfrac{25}{30}

\\ \tt\longmapsto Acceleration=0.8m/s^2

6 0
3 years ago
Hurry i have until 4 to turn in
Mariana [72]

Answer:

9miles average speed and 23 seconds

7 0
3 years ago
A World-class sprinter can reach a top speed of about 11.5 m/s in the first 18.0 m of a race. What is the average acceleration o
irina [24]

Answer

a = 3.674 m / s ^ 2


t = 3.13 s

Using the kinematic equations for the movement we have:


h(t) = P_{0} + Vot + \frac{1}{2}at ^ 2 (1)


V_{f} = V_{0} + at (2)


Where:


P_{0} = initial position


V_{0}} = initial velocity


a = acceleration


t = time in seconds


V_{f} = final speed


We know:


P_{0}=0


V_{0}= 0

h = 18 m


V_{f} = 11.5\frac{m}{s}

  So:

 From (2) we have that: t =\frac{V_{f}}{a}


t =\frac{11.5}{a}

From (1) we have to:


h (t) = 0.5at ^ 2\\h = 18 = 0.5at ^ 2

Then we clear "a" to find the acceleration.


\frac{36}{t^2} = a\\a = \frac{36}{(\frac{11.5}{a})^2} \\\\a =\frac{11.5^2}{36}\\a = 3.674 m / s ^2

Then, the time it takes to reach this speed is:


t =\frac{V_{f}}{a}\\t =\frac{11.5}{3.674}\\t = 3.13 s

4 0
4 years ago
Read 2 more answers
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