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bulgar [2K]
2 years ago
6

Liquid X of volume 0.5m3 and density 900kgm-3 was mixed with liquid Y of volume 0.4m3 and density 800kgm-3. What was the density

of the mixture?​
Physics
1 answer:
masha68 [24]2 years ago
8 0

Answer:

Density of the mixture = 855.56kgm-3

Explanation:

Density = Mass / Volume

Volume of Liquid X = 0.5m³

Density of Liquid X = 900kgm-3

Mass of Liquid X = Density × Volume

= 900kgm-3 × 0.5m³ = 450kg

Volume of Liquid Y = 0.4m³

Density of Liquid Y = 800kgm-3

Mass of Liquid Y = Density × Volume

= 800kgm-3 × 0.4m³= 320kg

As X and Y are mixed, we add their masses and volumes together:

Mass = 770kg

Volume = 0.9m³

Now we can find the density of the mixture:

Density = 770kg / 0.9m³ = 855.56kgm-3 (rounded to the 2nd decimal)

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Which particles are located in the nucleus of the atom? (2 points)
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Neutrons and protons are located in the nucleus of the atom.

Explanation:

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3 years ago
A piece of metal has attained a velocity of 107.8 m/sec after fallinf for 10 seconds what is its initial velocity
soldi70 [24.7K]

Answer:

7.8 m/s

Explanation:

Here object is falling with a gravitational acceleration there  for we can take acceleration = 10 m/ s² and its constant through out the motion there for we can use motion equation

V = U + at

V - Final velocity

U - Initial velocity

a - acceleration

t - time

V=U+at

107.8=U + 10×10

  = 7.8 m/s

4 0
3 years ago
When bouncing a ball, the bouncing motion results in the ball ____________.
Alekssandra [29.7K]

Answer: "B" Changing Position

Great Question!

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8 0
3 years ago
Water is pumped steadily out of a flooded basement at a speed of 5.4 m/s through a uniform hose of radius 0.83 cm. The hose pass
Gala2k [10]

To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.

The work done would be defined as

\Delta W = \Delta PE + \Delta KE

Where,

PE = Potential Energy

KE = Kinetic Energy

\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2

Where,

m = Mass

g = Gravitational energy

h = Height

v = Velocity

Considering power as the change of energy as a function of time we will then have to

P = \frac{\Delta W}{\Delta t}

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

The rate of mass flow is,

\frac{\Delta m}{\Delta t} = \rho_w Av

Where,

\rho_w = Density of water

A = Area of the hose \rightarrow A=\pi r^2

The given radius is 0.83cm or 0.83 * 10^{-2}m, so the Area would be

A = \pi (0.83*10^{-2})^2

A = 0.0002164m^2

We have then that,

\frac{\Delta m}{\Delta t} = \rho_w Av

\frac{\Delta m}{\Delta t} = (1000)(0.0002164)(5.4)

\frac{\Delta m}{\Delta t} = 1.16856kg/s

Final the power of the pump would be,

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

P = (1.16856)((9.8)(3.5)+\frac{1}{2}5.4^2)

P = 57.1192W

Therefore the power of the pump is 57.11W

6 0
3 years ago
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