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3 years ago

6

A spacecraft in outer space increases velocity by firing its rockets. How can hot gases escaping from its rockets change the vel

ocity of the craft when there is nothing in space for the gases to push against?

1 answer:

Elena-2011 [213]3 years ago

3 0

Answer:

Momentum is conserved.

Explanation:

-Momentum is conserved.

-By Newton's third law (For every action, there is an equal and opposite reaction.)-the change in momentum of gases in one direction must be balanced by an equal change in momentum of the spacecraft in the opposite direction.

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Subduction occurs because the sea floor is: more dense than the continental floor less dense than the continental floor more buo

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A)more dense than the continental floor

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3 years ago

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How do i figure out this question?

nikklg [1K]

Answer:

0.75 g/cm^3

Explanation:

The formula for density:

$\rho = \frac{m}{V}$

Where m is the mass and V is the volume.

So, we can substitute values for m and V:

$\rho = \frac{277}{370}\approx0.75$

Therefore, the density is 0.75 g/cm^3 (watch the units!)

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A mass of 0.5 kg hangs motionless from a vertical spring whose length is 1.10 m and whose unstretched length is 0.50 m. Next the

ser-zykov [4K]

Answer:

The maximum length during the motion is $L_{max} = 1.45m$

Explanation:

From the question we are told that

The mass is $m =0.5 kg$

The vertical spring length is $L = 1.10m$

The unstretched length is $L_{un} = 1.30m$

The initial speed is $v_i = 1.3m/s$

The new length of the spring $L_{new} = 1.30 m$

The spring constant k is mathematically represented as

$k = -\frac{F}{y}$

Where F is the force applied $= m * g = 0.5 * 9.8=4.9N$

y is the difference in weight which is $=1.10-0.50=0.6m$

The negative sign is because the displacement of the spring (i.e its extension occurs against the force F)

Now substituting values accordingly

$k = \frac{4.9}{0.6}$

$= 8.17 N/m$

The elastic potential energy is given as $E_{PE} = \frac{1}{2} k D^2$

where D is this the is the displacement

Since Energy is conserved the total elastic potential energy would be

$E_T = initial \ elastic\ potential \ energy + kinetic \ energy$

$E_T = \frac{1}{2} k D_{max}^2 = \frac{1}{2} k D^2 + \frac{1}{2} mv^2$

Substituting value accordingly

$\frac{1}{2} *8.17 *D_{max}^2 =\frac{1}{2} * 8.17*(1.30 - 0.50)^2 + \frac{1}{2} * 0.5 *1.30^2$

$4.085 * D_{max}^2 = 3.69$

$D^2_{max} = 0.9033$

$D_{max} = 0.950m$

So to obtain total length we would add the unstretched length

So we have

$L_{max} = 0.950 + 0.5 = 1.45m$

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3 years ago

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State the methods of nuclear fission waste disposal and the suitability of each method.

Alexeev081 [22]

<span>Most low-level radioactive waste (LLW) is typically sent to land-based disposal immediately following its packaging for long-term management. This means that for the majority (~90% by volume) of all of the waste types produced by nuclear technologies, a satisfactory disposal means has been developed and is being implemented around the world.
</span>

Radioactive wastes are stored so as to avoid any chance of radiation exposure to people, or any pollution.The radioactivity of the wastes decays with time, providing a strong incentive to store high-level waste for about 50 years before disposal.Disposal of low-level waste is straightforward and can be undertaken safely almost anywhere.Storage of used fuel is normally under water for at least five years and then often in dry storage.<span>Deep geological disposal is widely agreed to be the best solution for final disposal of the most radioactive waste produced.

</span>I suggest this site on this subject http://www.world-nuclear.org/information-library/nuclear-fuel-cycle/nuclear-wastes/storage-and-dispo...

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4 years ago

Reil [10]

Answer:

Yes

Explanation:

When an object has more mass it takes more gravity to keep it down therefore producing friction which in return reduces the amount of kinetic energy created. A change in an object's speed has an greater effect on its kinetic energy. than a change in its mass has, because kinetic energy is proportional to.

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4 years ago