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andre [41]
2 years ago
11

why is it important for scientist and engineer who plan uncrewed missions to accurately know the masses and the paths of the pla

nets the spacecraft will visit?
Physics
1 answer:
vodka [1.7K]2 years ago
8 0

Answer:

u may find your answer from this link-

https://www.nasa.gov/50th/50th_magazine/scientists.html

Explanation:

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How does your body’s flexibility naturally change as you age?
chubhunter [2.5K]
<span>Your flexibility decreases.  But if you exercise or stretch a few times a  week you can slow down the process </span>
7 0
3 years ago
Read 2 more answers
What is the specific heat of an object if it takes 1200 J to raise the temperature of a 20 kg object by 6 degrees C?
ch4aika [34]

Answer:

c=10\ J/kg^{\circ} C

Explanation:

Given that,

Heat required, Q = 1200 J

Mass of the object, m = 20 kg

The increase in temperature, \Delta T=6^{\circ} C

We need to find the specific heat of the object. The heat required to raise the temperature is given by :

Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{1200}{20\times 6}\\\\c=10\ J/kg^{\circ} C

So, the specific heat of the object is 10\ J/kg^{\circ} C.

5 0
3 years ago
What are the three ways of answering a scientific question
My name is Ann [436]

Answer:

Let's start by understanding what exactly a scientific question is. A scientific question is a question that may lead to a hypothesis and help us in answering (or figuring out) the reason for some observation. A good scientific question has certain characteristics. It should have some answers (real answers), should be testable.

Here's examples of a few:

Why is that a star?

or

What is that star made of?

Hope this can lead you to the answer you're looking for at least!!

5 0
3 years ago
How long does it take to fall from 5000 feet?
WINSTONCH [101]
It depends on your weight, your hieght, and how fast you are falling
4 0
3 years ago
A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

7 0
3 years ago
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