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JulijaS [17]
2 years ago
7

QC In ideal flow, a liquid of density 850 kg / m³ moves from a horizontal tube of radius 1.00cm into a second horizontal tube of

radius 0.500cm at the same elevation as the first tube. The pressure differs by ΔP between the liquid in one tube and the liquid in the second tube. (b) ΔP = 6.00kPa and
Physics
1 answer:
kvv77 [185]2 years ago
4 0

The  volume flow rates for ∆P is 6.81m³/s .

<h3>What is pressure?</h3>

The amount of force applied on perpendicular to the surface of an object per unit area. The unit of it is pascal.

According to bernaulli's theorem theorem

P+1/2pV²+pgy = constant

where p fluid density

g is acceleration due to gravity, pressure at elevation,v is Velocity at elevation ,y is height of elevation.

As there are two tubes then the height of tube 1 is equal to height of tube two .

P1-P2=1/2p(Vd²-Vl²)

The flow rate of liquid is  A1V1=A2V2 .

rest is attached in image

to learn more about Pressure click here brainly.com/question/12971272

#SPJ4

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Answer:

4 m/s

Explanation:

Momentum is defined as:

p=mv

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m is the mass of the object

v is its velocity

For the object in this problem, we know:

p = 200 kg m/s is the momentum

m = 50 kg is the mass

Solving for the velocity, we find:

v=\frac{p}{m}=\frac{200}{50}=4 m/s

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4. A car accelerates, from rest, at 2.4 m/s?. How fast is the car traveling
pantera1 [17]

Answer:

19.2m/s

Explanation:

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3 years ago
. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it
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Answer:

29.4 N/m

0.1  

Explanation:

a) From the restoring Force we know that :  

F_r = —k*x  

the gravitational force :  

F_g=mg  

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force  

xi , x2 are the displacement made by the two masses.

Givens:

<em>m1 = 1.29 kg</em>

<em>m2 = 0.3 kg  </em>

<em>x1   = -0.75 m  </em>

<em>x2 = -0.2 m </em>

<em>g   = 9.8 m/s^2  </em>

Plugging known information to get :

F_r =F_g

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b) To get the unloaded length 1:  

l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m  

Plugging known infromation to get :

l= x1 — (F_1/k)  

= 0.1  

 

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3 years ago
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