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Yuki888 [10]
3 years ago
5

Un montacargas hidráulico es utilizado para levantar cajas con un peso de 1500 newtons sobre una superficie de 20 cm al cuadrado

. ¿Cuál es la fuerza que se aplica sobre una base de 2 cm al cuadrado?​
Physics
1 answer:
saveliy_v [14]3 years ago
3 0

Answer:

150 N

Explanation:

The question says that, "A hydraulic forklift is used to lift boxes with a weight of 1500 newtons over a surface of 20 cm squared. What is the force applied to a base of 2 cm squared?"

Given that,

Force applied, F₁ = 1500 N

Area, A₁ = 20 cm²

We need to find the force applied to a base of 2 cm². Using Pascal's law to find it such that,

\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}\\\\F_2=\dfrac{F_1A_2}{A_1}\\\\F_2=\dfrac{1500\times 2}{20}\\\\F_2=150\ N

So, the required weight is 150 N.

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An apple is placed 20.0 cm in front of a diverging lens of focal length 10.0 cm. Find the image distance and the magnification o
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Answer:

Image distance of apple=-6.7 cm

Magnification of apple=0.33

Explanation:

We are given that an apple is placed 20.cm in front of a diverging lens.

Object distance=u=-20 cm

Focal length=f=-10 cm

Because focal length of diverging lens is negative.

We have to find the image distance and magnification of the apple.

Lens formula

\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

Substitute the values then we get

-\frac{1}{10}=\frac{1}{v}+\frac{1}{20}

\frac{1}{v}=-\frac{1}{10}-\frac{1}{20}

\frac{1}{v}=\frac{-2-1}{20}=-\frac{3}{20}

v=-\frac{20}{3}=-6.7 cm

Image distance of apple=-6.7 cm

Magnification=m=\frac{v}{u}=\frac{-\frac{20}{3}}{-20}

Magnification of apple=\frac{1}{3}=0.33

Hence, the magnification of apple=0.33

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A spherical submersible 1.84 m in radius, armed with multiple cameras, descends under water in a region of the Atlantic Ocean kn
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Answer:

23932242.5 Pa

Explanation:

P_a = Atmospheric pressure = 1.013\times 10^5\ Pa

P_w = Pressure of seawater

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h = Depth of shipwreck = 2.37\times 10^3\ m

g = Acceleration due to gravity = 9.81 m/s²

The absolute pressure is given by

P_{ab}=P_a+P_w\\\Rightarrow P_{ab}=1.013\times 10^5+1.025\times 10^3\times 9.81\times 2.37\times 10^3\\\Rightarrow P_{ab}=23932242.5\ Pa

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