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Pie
2 years ago
15

Why is water not suitable for hydraulic machine​

Physics
1 answer:
Marina86 [1]2 years ago
4 0

Answer:

Hydraulic fluid has a higher boiling point than water to help combat this. Related to this is the concept of vapor pressure. Hydraulic systems often involve small orifices, which can cause cavitation (localized boiling). Water will also exacerbate galvanic corrosion when dissimilar metals are used in the system.

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You hold a bucket in one hand. In the bucket is a 500 g rock. You swing the bucket so the rock moves in a vertical circle 2.2 m
slavikrds [6]

Answer:v=3.28 m/s

Explanation:

Given

mass of rock m=500 gm

diameter of circle d=2.2 m

radius r=\frac{2.2}{2}=1.1 m

At highest Point

mg+N=\frac{mv^2}{r}

At highest Point N=0 because mass is just balanced by centripetal Force

thus mg=\frac{mv^2}{r}

v=\sqrt{gr}

v=\sqrt{9.8\times 1.1}

v=\sqrt{10.78}

v=3.28 m/s

6 0
3 years ago
4. Derive<br>the relation, P= hd g​
Sergeu [11.5K]

p=F/A

or,P=d×V×G/A (m=d×V)

or,p= d× A×h×g/A (A and A are cut)

or,P=d×H×G

4 0
3 years ago
What is capacitance?
just olya [345]

the amount of charge stored per volt

3 0
3 years ago
Sarah, whose mass is 40 kg, is on her way to school after a winter storm when she accidentally slips on a patch of ice whose coe
RideAnS [48]

Sarah's acceleration is -0.49 m/s^2

Explanation:

The force of kinetic friction acting on Sarah has a magnitude which is given by:

F_f = \mu mg

where

\mu is the coefficient of kinetic friction

m is Sarah's mass

g is the acceleration of gravity

Moreover, according to Newton's second law of motion, we know that the net force on Sarah is equal to its mass times its acceleration:

F=ma

where a is the acceleration

Since the force of friction is the only force acting on Sarah, we can say that the net force is equal to the force of friction, therefore:

F=-\mu mg = ma

where the negative sign is due to the fact that the force of friction has a direction opposite to the motion of Sarah. Solving for a, we find

a=-\mu g

And substituting the following values:

\mu = 0.05 (coefficient of friction)

g=9.81 m/s^2 (acceleration of gravity)

we find:

a=-(0.05)(9.81)=-0.49 m/s^2

Learn more about acceleration and forces:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

4 0
3 years ago
An airplane flying at an altitude of 6 miles passes directly over a radar antenna. When the airplane is 10 miles away (s = 10),
Novosadov [1.4K]

Answer:

Explanation:

Given

altitude of the Plane h=6\ miles

When Airplane is s=10\ miles away

Distance is changing at the rate of \frac{\mathrm{d} s}{\mathrm{d} t}=290\ mph

From diagram we can write as

h^2+x^2=s^2

differentiate above equation w.r.t time

2h\frac{\mathrm{d} h}{\mathrm{d} t}+2x\frac{\mathrm{d} x}{\mathrm{d} t}=2s\frac{\mathrm{d} s}{\mathrm{d} t}

as altitude is not changing therefore \frac{\mathrm{d} h}{\mathrm{d} t}=0

0+x\frac{\mathrm{d} x}{\mathrm{d} t}=s\frac{\mathrm{d} s}{\mathrm{d} t}

at s=10\ miles\ and\ h=6\ miles

substitute the value we get x=\sqrt{10^2-6^2}=8\ miles

8\times \frac{\mathrm{d} x}{\mathrm{d} t}=10\times 290

\frac{\mathrm{d} x}{\mathrm{d} t}=362.5\ mph

5 0
3 years ago
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