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2 years ago

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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?

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

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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