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bearhunter [10]
3 years ago
12

Why is the answer B? I'm not sure which equation to look at when doing these proportion types of problems.

Physics
1 answer:
loris [4]3 years ago
7 0

Explanation:

Given:

Δy = d

v₀ = 0

a = g

Find: v

v² = v₀² + 2aΔy

v² = (0)² + 2(g)(d)

v = √(2gd)

If we double d:

√(2g (2d))

√(4gd)

Compared to the first velocity:

√(4gd) / √(2gd)

√2

So the new velocity is √2 greater than the first velocity.

√2 v

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a certain hydraulic jack is used for lifting load. a force of 250N is applied on a small piston with a diameter of 80cm while th
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The maximum mass of a load that can be lifted by the jack and the distance covered are:

m = 160.2 Kg

h = 25 cm

Given that a certain hydraulic jack is used for lifting load. a force of 250N is applied on a small piston with a diameter of 80cm while the diameter of the larger piston is 2.0m.

The parameters given are

F_{1} = 250

A_{1} = Area of the small piston = πr^{2}

A_{1} = 22/7 x 0.4^{2}

A_{1} = 0.5 m^{2}

F_{2} = ?

A_{2} = Area of the large piston = πr^{2}

A_{2} = π x 1

A_{2} = 3.14 m^{2}

To calculate the force on the large piston, we will use the below formula

F_{1}/ A_{1} = F_{2} / A_{2}

Substitute all the parameters into the equation

250/0.5 =  F_{2}/3.14

F_{2} = 1570 N

To calculate the maximum mass of a load that can be lifted by the jack, let us apply Newton second law

F = mg

1570 = 9.8m

m = 1570/9.8

m = 160.2 Kg

.(take g=9.81ms^-2)​

If the applied force moves through a distance of 25cm, the distance through which the load is lifted will be

F_{1}/ 0.25A_{1} = F_{2} / A_{2}h

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make h the subject of the formula

6280h = 1570

h = 1570/6280

h = 0.25 m

Therefore, the distance through which the load is lifted is 25 cm

Learn more here: brainly.com/question/13596980

5 0
2 years ago
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