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stiks02 [169]
3 years ago
11

Where does the shadow of the shuttle in space form?

Physics
1 answer:
vekshin13 years ago
8 0
The shadow forms on the first surface away from the shuttle in the direction opposite the sun.
You might be interested in
usted / el partido de fútbol Usted va al partido de fútbol. Question 1 with 1 blankyo / la biblioteca Question 2 with 1 blanknos
34kurt

Answer:

1- Yo VOY A la biblioteca

2- Nosotros VAMOS A la piscina

3- Mis primas VAN AL gimnasio

4- Tú VAS de excursión

5- Ustedes VAN AL parque municipal

6- Alejandro VA AL museo de ciencias

Explanation:

Las preguntas refieren en su totalidad a la conjugación temporal del verbo "ir". Así, refiere a conjugar dicho verbo en tiempo presente simple, el cual se conjuga de la siguiente manera:

Yo voy, el va, tú vas, nosotros vamos, ellos van, ustedes van

Así, debe interpretarse la persona o personas que realizan la acción, para determinar así como se conjuga dicho verbo.

4 0
3 years ago
When jumping down from a table to the ground, it is helpful to bend your knees when you hit the ground because...
forsale [732]

Answer:

This reduces the average force applied during the landing process/ or you can say it reduces the impact your body takes.

Explanation:

7 0
2 years ago
Which quantity best describes this description? Justin walks at 3 miles per hour north, and then speeds up to 4 miles per hour.
prisoha [69]

Personally, I agree with your answer, namely that the likely-intended event happening here is one of acceleration. Having said that, I also want to add: it pains me to see this type of wording because, clearly, it is vague and only invites confusion of the type you are talking about.

Good luck!

4 0
3 years ago
If a moving object travels for a distance of 160 m. In 20 seconds what’s the average speed
Rus_ich [418]

Answer:

8.00

Explanation:

you have to divide

6 0
3 years ago
A long thin rod of mass M and length L is situated along the y axis with one end at the origin. A small spherical mass m1 is pla
Yuri [45]

Answer:

= - 5.65\times 10^{-2} J

Explanation:

Given data:

L =2.00 *10^4 m

d = 18*10^4 m

M = 18  *10^6 kg

m_1 = 8*10^6 kg

Gravitational energy is given as

U =- G \frac{m_1 m_2}{r}

mass per unit length is given as

\sigma = \frac{M}{L} = \frac{18 \times 10^6}{2\times 10^4 m} = 900 kg/m

calculating potential energy

dU ==-G\int_{16\times 10^4}^{18\times 10^4} \frac{m_1 *dm}{r}

=-G\int_{16\times 10^4}^{18\times 10^4} \frac{m_1 *\sigma dr}{r}

=-G*m_1*\sigma\int_{16\times 10^4}^{18\times 10^4} \frac{dr}{r}

=-G*m_1*\sigma \left | ln r \right |_{16\times 10^4}^{18\times 10^4}

=-G*m_1*\sigma ln(1.125)

=-6.673 \times 10^{-11}*8*10^6*900*ln(1.125)

= - 5.65\times 10^{-2} J

5 0
2 years ago
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