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3 years ago

Where should the fulcrum be located in this lever to get the greatest mechanical advantage?

2 answers:

8 0

Where should the fulcrum be located in this lever to get the greatest mechanical advantage?

Answer:C) at position I because the length of the effort arm should be greater than the length of the load arm

amid [387]3 years ago

4 0

C. When ever it comes to questions like these, I like to think of a seesaw with a huge amount of weight on one end. Which position would work best for that? With the effort arm being longer, you have more leverage so the load arm will move easier.

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You are given the melting points of three unknown substances and are asked to predict which one is an ionic compound. You would

vovangra [49]

To identify which one is ionic compound, you would select the compound with the high melting point. Ionic compounds have high melting points because the electrostatic interactions that hold the compounds together are very strong. Hope this answers the question.

8 0

3 years ago

Read 2 more answers

Will give brainiest 8th grade physics

Iteru [2.4K]

Answer:

Explanation:

Hope this helped!

7 0

3 years ago

300kg of water are lifted 10m vertically in 5s show the work done in 30kj and that power is 6kw . Please help me

Mandarinka [93]

Answer:

6KW

Explanation:

The computation is shown below:

We know that

Work done= m ×g× h

Here

W= 300×10×10

= 30000 J

= 30 KJ

And

Power= Work done ÷time taken

P = 30000 ÷ 5

= 6000W

= 6KW

The above represent the answer

3 0

3 years ago

Why is the earth's magnetic field important for us?

Nataly_w [17]

It protects us from the magnetic/electrical radiation that comes from the sun. High radiation periods coincide with solar storms.

8 0

3 years ago

Particle’s position along the x-axis is described by the function x(t) = A t + B t2,

sleet_krkn [62]

Answer

given,

x(t) = A t + B t²

A = -4.1 m/s

B = 5.4 m/s²

a) velocity of the particle is equal to the differentiation of Position w.r.t. time

$\dfrac{dx}{dt}=\dfrac{d}{dt}(-4.1 t + 5.4 t^2)$

$v =-4.1 + 2\times 5.4 t$

$v = -4.1 + 10.8 t$

the above equation gives the function of velocity.

b) time at which velocity of particle is zero

v = -4.1 + 10.8 t

inserting v = 0 and calculating v

0 = -4.1 + 10.8 t

10.8 t = 4.1

t = 0.38 s

time at which velocity is zero is equal to 0.38 s.

7 0

3 years ago