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Alex73 [517]
3 years ago
6

Express the volume expansivity and the isothermal compressibility as functions of density rho and its partial derivatives. For w

ater at 50°C and 1 bar, κ = 44.18 × 10 −6 bar −1 . To what pressure must water be compressed at 50°C to change its density by 1%? Assume that κ is independent of P.
Physics
1 answer:
kap26 [50]3 years ago
7 0

Answer:

Explanation:

κ = 1/v  x dv/dp

= 1/ρ x (+δρ/ δp )_T

δρ = 1 , ρ = 100

44.18 x 10⁻⁶ = ( 1/100) x( 1 / δp)

δp = 10⁻² / (44.18 x 10⁻⁶ )

= .0226 x 10⁴

= 226 bar

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A launched hopper reach to 1.20 m maximum height. How much is it’s launch velocity?
garri49 [273]

The launch velocity is 4.8 m/s

Explanation:

We can solve this problem by applying the law of conservation of energy. In fact, the mechanical energy of the hopper (equal to the sum of the potential energy + the kinetic energy) is conserved. So we can write:

U_i +K_i = U_f + K_f

where:

U_i is the initial potential energy, at the bottom

K_i is the initial kinetic energy, at the bottom

U_f is the final potential energy, at the top

K_f is the final kinetic energy, at the top

We can rewrite the equation as:

mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2

where:

m is the mass of the hopper

g=9.8 m/s^2 is the acceleration of gravity

h_i = 0 is the initial height

u is the launch speed of the hopper

h_f = 1.20 m is the maximum altitude reached by the hopper

v = 0 is the final speed (which is zero when the hopper reaches the maximum height)

Solving the equation for u, we find the launch speed of the hopper:

u=\sqrt{2gh_g}=\sqrt{2(9.8)(1.20)}=4.8 m/s

Learn more about kinetic energy and potential energy:

brainly.com/question/6536722

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brainly.com/question/10770261  

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4 0
3 years ago
Can someone help me with this please
andrezito [222]
Carbon: C, 12.011, 6, 12
Oxygen: O, 8, 8, 8, 16
Boron: B, 10.811, 5, 5, 11
3 0
2 years ago
A 9,100​-microfarad ​[mu​F] capacitor is charged to 22 volts​ [V]. If the capacitor is completely discharged through an iron rod
MakcuM [25]

Answer:

The temperature rises by 0.52K

Explanation:

Detailed explanation and calculation is shown in the image below

5 0
3 years ago
A man attaches a divider to an outdoor faucet so that water flows through a single pipe of radius 9.25 mm into four pipes, each
irinina [24]

Answer:

1.24 m/s

Explanation:

Metric unit conversion:

9.25 mm = 0.00925 m

5 mm = 0.005 m

The volume rate that flow through the single pipe is

\dot{V} = vA = 1.45 * \pi * 0.00925^2 = 0.00039 m^3/s

This volume rate should be constant and divided into the 4 narrower pipes, each of them would have a volume rate of

\dot{V_n} = \dot{V} / 4 = 0.00039 / 4 = 9.74\times10^{-5} m^3/s

So the flow speed of each of the narrower pipe is:

v_n = \frac{\dot{V_n}}{A_n} = \frac{\dot{V_n}}{\pi r_n^2}

v_n = \frac{9.74\times10^{-5}}{\pi 0.005^2} = 1.24 m/s

8 0
3 years ago
If the object represented by the FBD below has a mass of 2.5 kg, what is the acceleration of the object?
Debora [2.8K]

Answer:

4 m/s² down

Explanation:

We'll begin by calculating the net force acting on the object.

The net force acting on the object from the left and right side is zero because the same force is applied on both sides.

Next, we shall determine the net force acting on the object from the up and down side. This can be obtained as follow:

Force up (Fᵤ) = 15 N

Force down (Fₔ) = 25 N

Net force (Fₙ) =?

Fₙ = Fₔ – Fᵤ

Fₙ = 25 – 15

Fₙ = 10 N down

Finally, we shall determine the acceleration of the object. This can be obtained as follow:

Mass (ml= 2.5 Kg

Net force (Fₙ) = 10 N down

Acceleration (a) =?

Fₙ = ma

10 = 2.5 × a

Divide both side by 2.5

a = 10 / 2.5

a = 4 m/s² down

Therefore, the acceleration of the object is 4 m/s² down

6 0
3 years ago
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