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Alexxx [7]
3 years ago
5

What force must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2000-kg car (a large car) resti

ng on the slave cylinder

Physics
1 answer:
Tema [17]3 years ago
7 0

Explanation:

Below is an attachment containing the solution.

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Temperature is a measure of the average speed of the ____ of a thing.
Pie

Answer:

it is atoms and molecules

7 0
3 years ago
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An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
3 years ago
PLEASE HELP!!! WOULD REALLY APPRECIATE!
Eva8 [605]

Answer:

a. slope=rise/run

rise=0.02

run=-2

determined using the point (3,0.08) and (1,0.1) on the graph

slope=0.02/-2

= -0.01 or -1/100

b.area= area of trapizoid+ rectangle

((0.07+0.11)÷2)×4+1×0.07

0.36+0.07

=0.43$

c. the area represent the total cost after 5 hours

PLEASE MARK BRAINLIEST

3 0
2 years ago
What is the momentum of a 5.6 kg ball going 22m/s ?
maks197457 [2]
<h2>momentum(p) = mass(m)×acceleration (a)</h2>

<h3>p = 5.6 × 22</h3><h3>p = 123.2 kg m/s</h3>

6 0
3 years ago
In an experiment of a simple pendulum, measurements show that the pendulum has length し 0.397 ± 0.006 m, mass M-0.3172 ± 0.0002
vichka [17]

Answer:

1.)1.265+or minus 0.0006m

2).0.71%

Explanation:

See attached file

6 0
2 years ago
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