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marshall27 [118]
3 years ago
12

Once out in space, the space shuttle needs to ________________ to keep going in the same direction at the same speed.

Physics
2 answers:
laila [671]3 years ago
4 0
Let go one of its parts that is the boost
MissTica3 years ago
3 0
Accelerate.
Hope it helps<span />
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What is the significance of a standard system of measurement?
dimulka [17.4K]

Because scientists all over the world are working together, looking for answers to the same questions, just as much as if they all worked in the same physical laboratory in the same building.  They need a way to share data and experimental results in a form that everyone can understand. ( D )

Let's say I perform an experiment and get very exciting results. I'm a good scientist, so the next thing I want to do is to publish a complete description of how I did my experiment, and include all of my results.  That way, scientists around the world can read about what I did, they can find any mistakes that I made, and they can even repeat my experiment for themselves and see if they get the same results.

Now let's say that my results looked like this:

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The reaction stabilized when it reached the rate of 1.26 briligs per tove.

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After running at that constant rate for 35 toves, a pile of product was produced whose mass was exactly 61.284 wibbles.

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Do YOU understand my results ?

All those other scientists would have a tough time trying to decide whether my results made sense.  And if they repeated my experiment, they would have no way to tell whether their results matched mine or not.

Without a standard system of measurement, and units that mean the same thing to everybody, us scientists literally could not communicate.


3 0
2 years ago
Read 2 more answers
Topic Gravitational force amd firld strength.. help me please
I am Lyosha [343]

The gravitational force between <em>m₁</em> and <em>m₂</em> has magnitude

F_{1,2} = \dfrac{Gm_1m_2}{x^2}

while the gravitational force between <em>m₁</em> and <em>m₃</em> has magnitude

F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}

where <em>x</em> is measured in m.

The mass <em>m₁</em> is attracted to <em>m₂</em> in one direction, and attracted to <em>m₃</em> in the opposite direction such that <em>m₁</em> in equilibrium. So by Newton's second law, we have

F_{1,2} - F_{1,3} = 0

Solve for <em>x</em> :

\dfrac{Gm_1m_2}{x^2} = \dfrac{Gm_1m_3}{(15-x)^2} \\\\ \dfrac{m_2}{x^2} = \dfrac{m_3}{(15-x)^2} \\\\ \dfrac{(15-x)^2}{x^2} = \dfrac{m_3}{m_2} = \dfrac{60\,\rm kg}{40\,\rm kg} = \dfrac32 \\\\ \left(\dfrac{15-x}x\right)^2 = \dfrac32 \\\\ \left(\dfrac{15}x-1\right)^2 = \dfrac32 \\\\ \dfrac{15}x - 1 = \pm \sqrt{\dfrac32} \\\\ \dfrac{15}x = 1 \pm \sqrt{\dfrac32} \\\\ x = \dfrac{15}{1\pm\sqrt{\dfrac32}}

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

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5 0
2 years ago
Two parallel copper rods supply power to a high-energy experiment, carrying the same current in opposite directions. The rods ar
mart [117]

Answer:

the maximum allowable current is 7302.967  amperl

Explanation:

The computation of the maximum allowable current is shown below;

Force F = mean ÷ 4π 2 I_1 I_2 ÷d  × ΔL

200 N = (10)^-7 (2I × I) ÷ 0.08 × 1.5

200 = 3.75 × 10^-6 I^2

I = √200 ÷ √ 3.75 × 10^-6

= 7302.967  amperl

Hence, the maximum allowable current is 7302.967  amperl

Basically we applied the above formula

6 0
2 years ago
Why does fulcrum placement matter for how a lever works?
spin [16.1K]

Answer:

The mechanical advantage of using a lever is affected by the distance between the effort and the fulcrum and by the placement of the load. ... When the fulcrum is centered between the load and the lift, the amount of effort exerted to push down on the lever equals the amount of the load being lifted on the other end.

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3 years ago
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Damm [24]

Answer:

Stayed the same

Explanation:

Potential energy is a function of mass, gravity and height relative to a reference level. If we take as the reference level the soil, this is the level where the potential energy is zero. Since in problem it is mentioned that the track is flat, this means that there are no height changes with respect to the reference level, therefore we can say that the potential energy remains unchanged.

Ep = m*g*h

where:

m = mass [kg]

g = gravity [m/s^2]

h = elevation [m]

6 0
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