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ololo11 [35]
3 years ago
15

How is artificial gravity created in spaceships and sattelites ?????? U need some IQ for this

Physics
2 answers:
abruzzese [7]3 years ago
5 0

Artificial gravity can be created using a centripetal force. A centripetal force directed towards the centre of the turn is required for any object to move in a circular path.

goldfiish [28.3K]3 years ago
4 0

Explanation:

Artificial gravity can be created using a centripetal force. A centripetal force directed towards the center of the turn is required for any object to move in a circular path. In the context of a rotating space station it is the normal force provided by the spacecraft's hull that acts as centripetal force.

Hope it helps.

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a new planet is found with a density one third as much at earth and a radius twice that of earth. what is the acceleration due t
IrinaVladis [17]

The acceleration due to gravity on the new planet would be one third that of Earth's, or about 9.8 m/s^2.

What is planet?
This question might seem to have an easy solution, but it doesn't. The planets Earth, Mars, and Jupiter are well known. But before recent discoveries sparked a passionate debate about how to best characterise them, both Pluto and Ceres have been thought of as planets. This debate is still going strong today. The International Astronomical Union adopted the most recent definition of the a planet in 2006.

It specifies three requirements for a planet:

-It must move about a star (in our cosmic neighborhood, the Sun).
-It must be large enough for gravity to pull it in a spherical direction.
-It must have been large enough that any nearby objects of the a similar size were removed by its gravitational pull.

Given that the mass of the new planet is one third that of Earth's and its radius is twice that of Earth's, we can say that its volume is eight times that of Earth's.  This means that its average density is one third that of Earth. Therefore, the acceleration due to gravity on the new planet would be one third that of Earth's, or about 9.8 m/s^2.

To learn more about planet
brainly.com/question/13106554
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8 0
1 year ago
What variables affect density weight,conductivity,color,volume, or mass
tatiyna

Answer:

Density is affected by volume and mass.

Explanation:

Density is defined as the quantity of mass per unit of volume, or expressed mathematically, d = m/v.

6 0
3 years ago
During the warm-up and your scheduled physical activity, what was the weather like? Did the
yulyashka [42]
Yes, if weather was hot you need more fluids in your body due to the sweat pores needing to cool you down.
5 0
3 years ago
(5) A 4 kg. object rests on a flat, horizontal surface with a static
baherus [9]

Answer:

F = 9.81 [N]

Explanation:

To solve this problem we must use Newton's third le which tells us that the sum of forces on a body that remains static must be equal to one resulting from these forces in the opposite direction.

Let's perform a summation of forces on the vertical axis-y to determine the normal force N.

∑F = 0 (axis-y)

N - m*g = 0

where:

m = mass = 4 [kg]

g = gravity acceleration = 9.81 [m/s²]

N - (4*9.81)=0\\N = 39.24 [N]

Now we know that the frictional force can be calculated using the following equation.

f = μ*N

where:

f = friction force [N]

μ = friction coefficient = 0.25

N = normal force = 39.24 [N]

Now replacing:

f = 0.25*39.24\\f = 9.81[N]

Then we perform a sum of forces on the X-axis equal to zero. This sum of forces allows us to determine the minimum force to be able to move the object in a horizontal direction.

∑F = 0 (axis-x)

F-f=0\\F-9.81=0\\F= 9.81[N]

If the coefficient was smaller, a smaller force (F) would be needed to start the movement, this can be easily seen by replacing the value of 0.25, by smaller values, such as 0.1 or 0.05.

If the coefficient were larger, a larger force would be needed.

3 0
3 years ago
When the plutonium bomb was tested in New Mexico in 1945, approximately 1 gram of matter was converted into energy. Suppose anot
gayaneshka [121]

Answer:

The value is E =  1.35 *10^{14} \ J

Explanation:

From the question we are told that

    The mass of matter converted to energy on first test is  m  =  1 \  g  = 0.001 \  kg

    The mass of matter converted to energy on second test m_1 =  1.5 \  g = 1.5 *10^{-3} \ kg

    Generally the amount of energy that was released by  the explosion is  mathematically  represented as  

         E =  m * c^2

=>       E =  1.5 *10^{-3}  * [ 3.0 *10^{8}]^2

=>       E =  1.35 *10^{14} \ J

7 0
3 years ago
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