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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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A 1000-kg whale swims horizontally to the right at a speed of 6.0 m/s. it suddenly collides directly with a stationary seal of m
Sphinxa [80]

M = mass of the whale = 1000 kg

m = mass of the seal = 200 kg

V = initial velocity of whale before collision with the seal = 6.0 m/s

v = initial velocity of the seal before collision with the whale = 0 m/s

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Using conservation of momentum

M V + m v = (M + m) V'

inserting the above values in the equation

(1000 kg) (6.0 m/s) + (200 kg) (0 m/s ) = (1000 kg + 200 kg) V'

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3 0
3 years ago
Read 2 more answers
A solid nonconducting sphere of radiusRcarries a chargeQdistributed uniformly throughout itsvolume. At a certain distancer1(r1&l
lara [203]

Answer:

E' = \frac{1}{8} E

Explanation:

Given data:

first case

Distance of electric field from center of sphere is r_1 <R

Electric field at r_1< R

E = \frac{kQr_1}{R^3}

second case

Distance of electric field from centre of sphere is r_1 < 2R

Electric field at r_1< 2R

E' = \frac{kQr_1}{8R^3}

so, we have

E' = \frac{1}{8} E

3 0
3 years ago
A solid conducting sphere (radius = 5.0 cm) has a charge of 0.25 nC distributed uniformly on its surface. If point A is located
nadezda [96]

Answer:

\Delta V = 30V

Explanation:

give data:

inside diameter = 5.0 cm

charge q = 0.25 nC

Outside diameter = 15 cm

potential V at inside sphere is = V=\frac{kq}{R}

potential V at outside sphere is = V=\frac{kq}{r}

k is constant whose value is = 8.99 *10^{9}Nm^{2}/c^2

then potential difference between two point is

\Delta V = kq \left [\frac{1}{R}-\frac{1}{r}  \right ]

\Delta V = 9*10^{9}*0.25*10^{-9} \left [\frac{1}{0.05}-\frac{1}{0.15} \right ]

\Delta V = 30V

6 0
3 years ago
A curved road has a radius of 120 m and a cant angle of 48 degrees. What is the maximum speed to stay on the curve in the absenc
zalisa [80]

Answer:

Maximum speed, v = 36 m/s                                                          

Explanation:

Given that,

The radius of the curved road, r = 120 m

Road is at an angle of 48 degrees. We need to find the maximum speed of stay on the curve in the absence of friction. On a banked curve, the angle at which it is cant is given by :

tan\theta=\dfrac{v^2}{rg}

g is the acceleration due to gravity  

v=\sqrt{rg\ tan\theta}

v=\sqrt{120\times 9.8\times \ tan(48)}

v = 36.13 m/s

or

v = 36 m/s

So, the maximum speed to stay on the curve in the absence of friction is 36 m/s. Hence, this is the required solution.

7 0
3 years ago
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