True or False. The fulcrum of a first class lever is between the output force and an input force.

Answers

Answer 1

Answer:

True

Explanation:

A first-class lever is a lever in which the fulcrum is located between the input effort and the output load. In operation, a force is applied (by pulling or pushing) to a section of the bar, which causes the lever to swing about the fulcrum, overcoming the resistance force on the opposite side

Related Questions

Write down the use of light pipe.???

Answers

Light pipes are used to illuminate small areas. For example, light from a small LED to a switch indicator or a small graphic. Typically, a light pipe would go from a circuit to the area that requires illumination.

I need some help with this table, i am unsure on how to answer it.

Answers

Answer:

I can't see it sorry I can't help

Answer:

i cant see it good could you resend it please thanks

how could you use Newton's Second Law to study forces?

Answers

Answer:

F = ma or force is equal to mass times acceleration.

Explanation:

Newton's second law of motion is F = ma, or force is equal to mass times acceleration. Learn how to use the formula to calculate acceleration.
mark brainlist please

An illustration of the major tectonic plates is provided below.
Which of the labeled plates is the North American Plate?
A
Plate A
O
B
Plate B
c
Plate C
Plate D

Answers

Answer:

Plate A

Explanation: It has the continent of North America inside of it.

Answer:

its A

Explanation:Promise I did this rn and got it right

When should you get a vaccine

Answers

Answer:

Explanation:

as soon as you are offered one :P

A police car is moving east at 20m/s towards a driver who is moving west at 25m/s. The police car emits a frequency of 900Hz. What frequency is detected by the driver? The speed of sound at the time of this occurrence was 343m/s.

Answers

Answer:

f = 2445 Hz

Explanation:

i did the quiz

The frequency that the driver (observer) detects is equal to 1008 Hz

Given the following data:

Observer velocity = 25.0 m/sFrequency of sound = 900 HzSource velocity = 20.0 m/s Speed of sound = 343 m/s

To calculate the frequency that the driver (observer) detects, we would apply Doppler's effect of sound waves:

Mathematically, Doppler's effect of sound waves is given by the formula:

[tex]F_o = \frac{V \;+ \;V_o}{V\; - \;V_s} F[/tex]

Where:

V is the speed of a sound wave.F is the actual frequency of sound.[tex]V_o[/tex] is the observer velocity.[tex]V_s[/tex] is the source velocity.[tex]F_o[/tex] is the observer frequency.

Substituting the given parameters into the formula, we have;

[tex]F_o = \frac{343 \;+ \;20}{343\; - \;20} \times 900\\\\F_o = \frac{363}{323} \times 900\\\\F_o = 1.12 \times 900\\\\F_o =1008 \;Hz[/tex]

Driver (observer) frequency = 1008 Hz

Read more: https://brainly.com/question/23460034

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