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3 years ago

Which of the following is a pure substance?

Physics

2 answers:

astraxan [27]3 years ago

7 0

Yes I agree that it is D as well

11Alexandr11 [23.1K]3 years ago

6 0

The answer to your question is D

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An object moves from point A to point C along the rectangle shown in the figure below.

Naily [24]

Answer:

Hello friend where is the figure of the question

4 0

2 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

Does anyone know what the answers are?

baherus [9]

Answer:

Option (C) is the answer

Explanation:

may be it is possible if that we stand so far

4 0

3 years ago

How does dance use energy to influence motion?

Goshia [24]

Solar system energy. electoral power engery motion force engery . wind energy. kinetic energy

8 0

3 years ago

Consider a particle with initial velocity v⃗ that has magnitude 12.0 m/s and is directed 60.0 degrees above the negative x axis.

kramer

Answer:

$v_x = -6\ m/s$

Explanation:

initial velocity

magnitude of velocity, v = 12 m/s

angle made of velocity with negative x-axis,θ = 60°

We need to calculate x- component of v

$v_x = v cos \theta$

velocity is in negative x-direction, v = -12 m/s

now,

$v_x = -12\times cos 60^0$

$v_x = -12\times 0.5$

$v_x = -6\ m/s$

Hence, the velocity x-component is equal to -6 m/s.

5 0

3 years ago