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krok68 [10]
4 years ago
7

A pure substance can be a

Physics
1 answer:
valina [46]4 years ago
4 0
Hey there! A pure substance can be a element.

Hope this helps! :)
You might be interested in
5. An ion consist of 12 electrons and 11 proton. What is the charge<br> on the ion?
sammy [17]

Answer:

-1

Explanation:

Electrons have a negative charge and protons have a positive charge. (+11) + (-12) = -1

5 0
3 years ago
(1 point) At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 22 knots and ship B is sailing nort
Veronika [31]

Answer:28.8 knots

Explanation:

The ships are moving as the sides of a right triangle. Thus, Pyhogorean theorem will be useful in the following steps. Next, we have to know that the rate of change in distance, which is called velocity, can be described in terms of derivatives.

First, we have to calculate the distances covered by the ships from noon to 6 PM. In 6 hours, ship A moved 22*6=132 nautical mile. However, their first distance was 10 nautical miles, so 132+10=142 miles is the equivalent of A's displacement. For B, the distance travelled is 19*6=114 miles. From now on, A=142 miles and B=114 miles.

The distance between them is described with Pythogorean theorem, which is D=\sqrt{A^{2} +B^{2} } and when we replace the values A and D, we find Distance (D) to be 182 miles.

Now, let's make the notations clear. The velocity of A and B is notated as \frac{dA}{dt} and \frac{dB}{dt}. The rate of change of distance is also notated as \frac{dD}{dt}. Now, we have to find \frac{dD}{dt} from the Pythogorean theorem. If we derive the Pythogorean expression D=\sqrt{A^{2} +B^{2} } , we would have:

\frac{dD}{dt} =\frac{1}{2} *(A^{2} +B^{2} )^{-1/2} *(2*A*\frac{dA}{dt} + 2*B*\frac{dB}{dt} )

The derivation here includes chain rule and derives the interior parts of the parenthesis. When we insert distances for A and B and velocities for derivation notations, the formula becomes:

\frac{dC}{dt} =\frac{1}{2}*(142^{2}   +114^{2})^{-\frac{1}{2} }*(2*142*22 + 2*114*19) and the answer is 28.6 knots.

6 0
3 years ago
Scientists would like to learn more about the seismic activity that occurs on Earth’s surface. They are set to take a journey to
Ludmilka [50]

Answer:

the questions they could ask would be:

how much will the temperature change?

how much pressure will there be?

how durable does the equipment need to be?

how do we know if we will run into a water pocket or a magma pocket?

Explanation:

3 0
3 years ago
a circular cylinder and isused to maintain a water depth of 4 m. That is, when the water depth exceeds 4 m, thegate opens slight
stiv31 [10]

Answer:

  W / A = 39200 kg / m²

Explanation:

For this problem let's use the equilibrium equation of / newton

           F = W

Where F is the force of the door and W the weight of water

         W = mg

We use the concept of density

        ρ = m / V

        m = ρ V

The volume of the water column is

          V = A h

We replace

         W = ρ A h g

On the other side the cylinder cover has a pressure

          P = F / A

          F = P A

We match the two equations

       P A = ρ A h g

        P = ρ g h

        P = 39200 Pa

The weight of the water column is

       W  = 1000 9.8 4 A

       W / A = 39200 kg / m²

3 0
4 years ago
LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.
Svetradugi [14.3K]

Answer:

i) E = 269 [MJ]    ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

8 0
3 years ago
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