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vova2212 [387]
3 years ago
8

To lift an objet weighing 21,000N, how much force is needed on a piston with an area of 0.060 meters squared if the platform bei

ng lifted has an area of 3.0 meters squared?
Physics
1 answer:
Nezavi [6.7K]3 years ago
8 0

Answer:

The weight required on the piston equals 420 Newtons.

Explanation:

from the principle of transmission of pressure in a hydraulic lift  we have

\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{1}}

where

'F' is the force that acts on the piston

'A' is the area of the piston.

Since the force in the question is 21000 Newtons thus upon putting the values in the above equation we get

\frac{21000}{3.0m^{2}}=\frac{W}{0.06m^{2}}

Solving for W we get

W=\frac{21000}{3}\times 0.06\\\\\therefore W=420Newtons

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Rama09 [41]
I found this using the app Socratic. When I took physics in high school it helped me so much.

5 0
3 years ago
If pressure is increased from 200 kPa to 300 kPa, and the original volume of gas was 1.5 L, what is the new volume? Assume the t
Oxana [17]

Answer:

The answer to your question is:      V2 = 1 l

Explanation:

Data

P1 = 200 kPa

P2 = 300 kPa

V1 = 1.5 l

V2 = ?

Formula

                          P1V1 = P2V2

                          V2 = (P1V1) / P2

                          V2 = (200 x 1.5) / 300

                          V2 = 1 l

6 0
3 years ago
What is the magnitude of the sum of the two vectors A = 36 units at 53 degrees, and B =47 units at 157 degrees.
Thepotemich [5.8K]

Answer:

51.82

Explanation:

First of all, let's convert both vectors to cartesian coordinates:

Va = 36 < 53° = (36*cos(53), 36*sin(53))

Va = (21.67, 28.75)

Vb = 47 < 157° = (47*cos(157), 47*sin(157))

Vb = (-43.26, 18.36)

The sum of both vectors will be:

Va+Vb = (-21.59, 47.11)   Now we will calculate the module of this vector:

|Va+Vb| = \sqrt{(-21.59)^2+(47.11)^2}=51.82

4 0
3 years ago
Which landform is produced at location E where the Mississippi River enters the Gulf of
mixer [17]

Answer:

a delta

Explanation:

The landform produced at the location E where the Mississippi River enters the Gulf of Mexico is a delta.

A delta is a depositional landform where a smaller body of water enters into a larger one.

The Gulf of Mexico contains a larger body of water and as the Mississippi river enters into it, it splits up into many distributaries.

So, this feature is a delta.  

3 0
2 years ago
A wheel 1.0 m in radius rotates with an angular acceleration of 4.0rad/s2 . (a) If the wheel’s initial angular velocity is 2.0 r
Oliga [24]

Answer:

(a) ωf= 42 rad/s

(b) θ = 220 rad

(c) at = 4 m/s²  ,  v = 42 m/s

Explanation:

The uniformly accelerated circular movement,  is a circular path movement in which the angular acceleration is constant.

There is tangential acceleration (at ) and is constant.

We apply the equations of circular motion uniformly accelerated :

ωf= ω₀ + α*t  Formula (1)

θ=  ω₀*t + (1/2)*α*t²  Formula (2)

at = α*R  Formula (3)

v= ω*R  Formula (4)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular velocity ( rad/s)

ωf: final angular velocity ( rad/s)

R : radius of the circular path (cm)

at : tangential acceleration (m/s²)

v : tangential speed (m/s)

Data

α = 4.0 rad/s² : wheel’s angular acceleration

t = 10 s

ω₀ = 2.0 rad/s  : wheel’s initial angular velocity

R = 1.0 m  : wheel’s radium

(a)  Wheel’s angular velocity after 10 s

We replace data in the formula (1):

ωf= ω₀ + α*t

ωf= 2 + (4)*(10)

ωf= 42 rad/s

(b) Angle that rotates the wheel in the 10 s interval

We replace data in the formula (2):

θ=  ω₀*t + (1/2)*α*t²

θ=  (2)*(10) + (1/2)*(4)*(10)²

θ=  220 rad  

θ=  220 rad  

(c) Tangential speed and acceleration of a point on the rim of the wheel at the end of the 10-s interval

We replace data in the Formula (3)

at = α*R = (4)(1)

at = 4 m/s²

We replace data in the Formula (4)

v= ω*R = (42)*(1)

v = 42 m/s

6 0
3 years ago
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