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2 years ago

Describe each class of lever and explain to characteristics of each

Physics

2 answers:

Nataly [62]2 years ago

8 0

-- Class I lever

The fulcrum is between the effort and the load.

The Mechanical Advantage can be anything, more or less than 1 .

Example: a see-saw

-- Class II lever

The load is between the fulcrum and the effort.

The Mechanical Advantage is always greater than 1 .

Example: a nut-cracker, a garlic press

-- Class III lever

The effort is between the fulcrum and the load.

The Mechanical Advantage is always less than 1 .

I can't think of an example right now.

harina [27]2 years ago

6 0

Answer:

First Class of Lever: In this, Fulcrum is always changes the direction of the input force and can be used to increase the force or the distance. Second Class of Lever: In this, Fulcrum does not change direction of the input force & Output force is greater than the input force. Third Class of Lever: In this, the input force is between the fulcrum and the load does not change the direction of the input force. Here, Output force is less than input force.

Explanation:

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Can a small child play with fat child on the seesaw?Explain how?

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Yes a small child can play with fat child in the seesaw because if the the load distance is decreased the effort will increase. That's means if the distance between the fat boy and the fulcrum is decreased the small child needs less effort.so,he can play

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2 years ago

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Consider the following spectrum where two colorful lines (A and B) are positioned on a dark background. The violet end of the sp

Dmitry_Shevchenko [17]

Answer:

Explanation:

a )

This type of spectrum is called line emission spectrum . Because it consists of lines . It is emission spectrum because it is due to emission of radiation from a source .

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1 year ago

In rural areas, water is often extracted from underground by pumps. Consider an underground water source whose free surface is 6

stealth61 [152]

Answer:

W = 9533.09 Watt

Explanation:

given,

diameter of pipe inlet, d₁ = 10 cm

r₁ = 5 cm

diameter of pipe outlet, d₂ = 15 cm

r₂= 7.5 cm

head upto water level is to rise = 60 + 5

= 65 m

flow rate = 0.015 m³/s

we know

A₁ v₁ = A₂ v₂ = Q

π r₁² v₁ = π r₂² v₂ = 0.015

$v_1= \dfrac{r_2^2}{r_1^2} v_2$

$v_1= \dfrac{7.5^2}{5^2} v_2$

$v_1= 2.25 v_2$

$v_2 = \dfrac{0.015}{\pi r_2^2}$

$v_2 = \dfrac{0.015}{\pi 0.075^2}$

v₂ = 0.848 m/s

v₁ = 1.908 m/s

Applying Bernoulli's equation

$P_p = \dfrac{1}{2}\rho (v_2^2-v_1^2)+ \rho g h$

$P_p= \dfrac{1}{2}\times 1000\times (0.848^2-1.908^2)+ 1000\times 9.8\times 65$

$P_p= 635539.32 Pa$

P_p is the pump pressure

Power of the pump

W = P_p x Q

W = 635539.32 x 0.015

W = 9533.09 Watt

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2 years ago

If no friction acts on a diver during a dive, then which of the following statements is true? A) The total mechanical energy of

EleoNora [17]

If no frictional work is considered, then the energy of the system (the driver at all positions is conserved.

Let
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position 2 = final height of the diver (h₂) and the final velocity (v₂).

The initial PE = mgh₁ and the initial KE = (1/2)mv₁²
where g = acceleration due to gravity,
m = mass of the diver.
Similarly, the final PE and KE are respectively mgh₂ and (1/2)mv₂².
PE in position 1 is converted into KE due to the loss in height from position 1 to position 2.

Therefore
(KE + PE) ₁ = (KE + PE)₂

Evaluate the given answers.
A) The total mechanical energy of the system increases.
FALSE

B) Potential energy can be converted into kinetic energy but not vice versa.
TRUE

C) (KE + PE)beginning = (KE + PE) end.
TRUE

D) All of the above.
FALSE

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