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olga2289 [7]
3 years ago
7

A hydraulic lift is made by sealing an ideal fluid inside a container with an input piston of cross-sectional area 0.004 m2 , an

d an output piston of cross-sectional area 1.2 m2 . The pistons can slide up or down without friction while keeping the fluid sealed inside. What is the maximum weight that can be lifted when a force of 60 N is applied to the input piston
Physics
1 answer:
kvv77 [185]3 years ago
3 0

Answer:

Maximum weight that can be lifted = 18,000 N

Explanation:

Given:

Cross-sectional area of input (A1) = 0.004 m²

Cross-sectional area of the output (A2) = 1.2 m ²

Force (F) = 60 N

Computation:

Pressure on input piston (P1) = F / A1

Assume,

Maximum weight lifted by piston = W

Pressure on output piston (P2) = W / A2

We, know that

P1 = P2

[F / A1]  = [W / A2]

[60 / 0.004] = [W / 1.2]

150,00 = W / 1.2

Weight = 18,000 N

Maximum weight that can be lifted = 18,000 N

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⦁ A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal. (a)
GREYUIT [131]

Answer:

303.29N and 1.44m/s^2

Explanation:

Make sure to label each vector with none, mg, fk, a, FN or T

Given

Mass m = 68.0 kg

Angle θ = 15.0°

g = 9.8m/s^2

Coefficient of static friction μs = 0.50

Coefficient of kinetic friction μk =0.35

Solution

Vertically

N = mg - Fsinθ

Horizontally

Fs = F cos θ

μsN = Fcos θ

μs( mg- Fsinθ) = Fcos θ

μsmg - μsFsinθ = Fcos θ

μsmg = Fcos θ + μsFsinθ

F = μsmg/ cos θ + μs sinθ

F = 0.5×68×9.8/cos 15×0.5×sin15

F = 332.2/0.9659+0.5×0.2588

F =332.2/1.0953

F = 303.29N

Fnet = F - Fk

ma = F - μkN

a = F - μk( mg - Fsinθ)

a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0

303.29-0.35( 666.4 - 303.29*0.2588)/68.0

303.29-0.35(666.4-78.491)/68.0

303.29-0.35(587.90)/68.0

(303.29-205.45)/68.0

97.83/68.0

a = 1.438m/s^2

a = 1.44m/s^2

7 0
3 years ago
Formula:
s2008m [1.1K]

Answer:

55N

Explanation:

Using Newton's second law of motion:

F=ma

Force=mass × acceleration

F=25×2.2

F=55N

So 55 Newtons are needed

8 0
3 years ago
A hypothetical planet has a mass 2.81 times that of Earth, but the same radius.
patriot [66]

The acceleration due to gravity near the surface of the planet is 27.38 m/s².

<h3>Acceleration due to gravity near the surface of the planet</h3>

g = GM/R²

where;

  • G is universal gravitation constant
  • M is mass of the planet
  • R is radius of the planet
  • g is acceleration due to gravity = ?

g = (6.626 x 10⁻¹¹ x 2.81 x 5.97 x 10²⁴) / (6371 x 10³)²

g = 27.38 m/s²

Thus, the acceleration due to gravity near the surface of the planet is 27.38 m/s².

Learn more about acceleration due to gravity here: brainly.com/question/88039

#SPJ1

4 0
1 year ago
How much is the plane's acceleration while breaking if it takes 15 s for its velocity
Marizza181 [45]

Answer:

A. 4.67 m/s²

Explanation:

u = 145 m/s

v = 75 m/s

t = 15 s

a = v - u / t

= 145 - 75 / 15

= 4.67 m/s²

Hope this helped...

7 0
3 years ago
What type of friction occurs when you are trying to move an object, but the object isnt moving?
Tasya [4]

the answer is static friction


7 0
3 years ago
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