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Alla [95]
2 years ago
13

A dish is given to you, which contains a blackish-yellow powder. When you move a magnet over it, you are amazed to see black par

ticles, (which you find out are iron) fly upwards and get stuck to the magnet, and all that's left in the dish is a yellow powder, which you discover to be sulfur. Was your original powder an element, compound or mixture?
Physics
1 answer:
myrzilka [38]2 years ago
7 0

Answer:

In the case of the blackish-yellow powder, the black color is due to the presence of a iron and yellow color is due to sulfur. As the iron has retained its original properties it has got attached to the magnet. Thus the blackish-yellow powder is considered a mixture.

Explanation:

mark me brainliest

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A cork shoots out of a champagne bottle at an angle of 40.0 above the horizontal. If the cork travels a horizontal distance of 1
ruslelena [56]

Answer:

The initial speed of the cork was 1.57 m/s.

Explanation:

Hi there!

The equation of the horizontal position of the cork in function of time is the following:

x = x0 + v0 · t · cos θ

Where:

x = horizontal position at time t.

x0 = initial horizontal position.

v0 = initial speed of the cork.

t = time.

θ = launching angle.

If we place the origin of the frame of reference at the launching point, then x0 = 0.

We know that at t = 1.25 s, x = 1.50 m. We also know the launching angle so we can solve the equation of horizontal position for the initial speed, v0:

x = v0 · t · cos θ

x / t · cos θ = v0

v0 = 1.50 m / (1.25 s · cos (40.0°)

v0 = 1.57 m/s

The initial speed of the cork was 1.57 m/s.

4 0
3 years ago
Read 2 more answers
Examine the roller coaster track above. Assume there is negligible friction as the roller coaster moves from position A to posit
anyanavicka [17]

At point E

  • the kinetic energy of the rollercoaster is small compared to the potential energy
  • the potential energy is greater than the kinetic energy
  • the total energy is a mixture of potential and kinetic energy

<h3>What is the energy of the roller coaster at point E?</h3>

The energy of a roller coaster could either be potential energy, kinetic energy or a combination of both potential and kinetic energy.

Using analogies, the energy of the roller coaster at point E can be compared to a falling fruit from a tree which falls onto a pavement and is the rolling towards the floor. Point E can be compared to the midpoint of the fall of the fruit.

At point E

  • the kinetic energy of the rollercoaster is small compared to the potential energy
  • the potential energy is greater than the kinetic energy
  • the total energy is a mixture of potential and kinetic energy

In conclusion, the energy of the rollercoaster at E is both Kinetic and potential energy,

Learn more about potential and kinetic energy at: brainly.com/question/18963960

#SPJ1

5 0
2 years ago
Which of the following is not possible? A. Gas flow equals pressure gradient over resistance. B. Resistance equals pressure grad
kaheart [24]

Answer:

C. Pressure gradient equals gas flow over resistance.

Explanation:

As we know that pressure gradient is the driving force for the gas to flow from one point to other point

And we know that the flow rate is directly proportional to the driving force and it inversely depends on the resistance to flow

so we can say

Flow Rate = \frac{Driving \: force}{Resistance}

Flow Rate = \frac{Pressure \: Gradient}{Resistance}

so we can say that correct statements are as below

A. Gas flow equals pressure gradient over resistance.

B. Resistance equals pressure gradient over gas flow.

D. The amount of gas flowing in and out of the alveoli is directly proportional to the difference in pressure or pressure gradient between the external atmosphere and the alveoli.

5 0
3 years ago
PLZ HELP ASAP WILL GIVE BRAINLIEST!!!
igomit [66]

the answer is the forth one treatment of cancer

3 0
3 years ago
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Bonus: (It's not that hard, you just have to pay attention to units.) The Saturn V rocket first stage
agasfer [191]

v = 2.45×10^3\:\text{m/s}

Explanation:

Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as

F = \dfrac{d{p}}{d{t}} (1)

Assuming that the velocity remains constant then

F = \dfrac{d}{dt}(mv) = v\dfrac{dm}{dt}

Solving for v, we get

v = \dfrac{F}{\left(\frac{dm}{dt}\right)}\;\;\;\;\;\;\;(2)

Before we plug in the given values, we need to convert them first to their appropriate units:

The thrust <em>F</em><em> </em> is

F = 7.5×10^6\:\text{lbs}×\dfrac{4.45\:\text{N}}{1\:\text{lb}} = 3.34×10^7\:\text{N}

The exhaust rate dm/dt is

\dfrac{dm}{dt} = 15\dfrac{T}{s}×\dfrac{2000\:\text{lbs}}{1\:\text{T}}×\dfrac{1\:\text{kg}}{2.2\:\text{lbs}}

\;\;\;\;\;= 1.36×10^4\:\text{kg/s}

Therefore, the velocity at which the exhaust gases exit the engines is

v = \dfrac{F}{\left(\frac{dm}{dt}\right)} = \dfrac{3.34×10^7\:\text{N}}{1.36×10^4\:\text{kg/s}}

\;\;\;= 2.45×10^3\:\text{m/s}

6 0
2 years ago
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