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NeX [460]
3 years ago
15

When rockets split do they come back to earth?

Physics
1 answer:
attashe74 [19]3 years ago
4 0
I'm not 100% sure, but I believe what you mean is when they eject the old propulsion motors.  Yes, they land in the ocean and the US Navy retrieves them for later use.
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Can someone please help for a , what am I meant to do
monitta
I’m not sure what your question is but I hope I can be of help. Series would split voltage and have same current all through, whereas parallel would have the same voltage all throughout but would have different amperes(currents)
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The total of 8 and 13
Stella [2.4K]
The total is 21. If you use a calculator or count with your fingers it can help make the answer easier so in the end you will get 21 as your total.


7 0
3 years ago
Read 2 more answers
A crew team rows a boat at a rate of 20 km/h in still water. Beginning at time = 0 minutes, the team rows for 30 minutes up a ri
umka21 [38]
Since they are rowing against the current, their velocity realtive to the groud is 20-1.5 = 18.5km/h
So the distance travelled is 18.5km/h*0.5h = 9.25km
Going back with the current their ground realative velocity is 20+1.5 = 21.5km/h so the time it takes them to return 9.25 km is 9.25km/(21.5km/h) = 0.43h ~= 26 min
7 0
3 years ago
Find the components to write this vector in unit vector notation: 63.5 A ​please help
IrinaVladis [17]

Vector is perpendicular to x axis or i component.

Hence i component is 0

j component is 63.5

\\ \sf\longmapsto \overrightharpoon{A}=0\hat{i} +63.5\hat{j}

6 0
3 years ago
A 70.0-kg skier is sliding at 4 m/s when they slide down a 2m high hill. At the bottom of the hill they run into a large 2800 N/
Katarina [22]

Answer:

\Delta x=245\ mm

Explanation:

Given:

  • mass of skier, m=70\ kg
  • initial velocity of skier, u=4\ m.s^{-1}
  • height of the hill, h=2\ m
  • spring constant, k=2800\ N.m^{-1}

<u>final velocity of skier before coming in contact of spring:</u>

Using eq. of motion:

v^2=u^2+gh

v^2=4^2+9.8\times 2

v=5.9666\ m.s^{-1}

<u>Now the time taken by the skier to reach down:</u>

v=u+gt

5.9666=4+9.8\ t

t=0.2007\ s

<u>Now we calculate force using Newton's second law:</u>

F=\frac{dp}{dt}

F=\frac{m(v-u)}{t}

F=\frac{70\times(5.9666-4)}{0.2007}

F\approx686\ N

<u>∴Compression in spring before the skier momentarily comes to rest:</u>

\Delta x=\frac{F}{k}

\Delta x=\frac{686}{2800}

\Delta x=0.245\ m

\Delta x=245\ mm

4 0
3 years ago
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