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3 years ago

Which spacecraft carried the first Spacelab? -Columbia -Challenger -Discovery -Atlantis

Physics

1 answer:

Nat2105 [25]3 years ago

4 0

The Atlantis spacecraft carried the first space lab.

Answer: Option 4

<u>Explanation: </u>

The first space lab is named as ATLAS 1 which is the abbreviation of Atmospheric Laboratory for Applications and Science. It is a short space lab set up in space to observe the atmospheric changes and other scientific experiments in the outer atmosphere of Earth from space.

It contains hi-tech instruments and facilities. It was a part of Phase I of NASA’s mission to planet Earth. This helped in better understanding of Earth’s outer and inner atmosphere. So, the spacecraft used to carry the ATLAS 1 is named as Atlantis.

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The flow of cold dense air under the influence of gravity is called a _________ .

Yuki888 [10]

Answer: Katabatic Wind.

If you have any further questions, please comment. Have a great day. Thanks!
:)

6 0

3 years ago

The coefficient of linear expansion of steel is 12 × 10-6 K-1 . What is the change in length of a 25-m steel bridge span when it

Westkost [7]

Answer:

0.012-m

Explanation:

∆L = α × Lo × (T-To)

α is the coefficient of linear expansion = 12 × 10-6 K-1

Lo = Initial length = 25-m

∆L = Change in length

(T-To) = 40 K

∆L = 12 × 10-6 × 25 × 40

∆L = 0.012-m

4 0

3 years ago

The electric flux through a square-shaped area of side 5 cm near a very large, thin, uniformly-charged sheet is found to be 3\ti

deff fn [24]

Answer:

Explanation:

Given

side of square shape $a=5\ cm$

Electric flux $\phi =3\times 10^{-5}\ N.m^2/C$

Permittivity of free space $\epsilon_0=8.85\times 10^{-12} \frac{C^2}{N.m^2}$

Flux is given by

$\phi =EA\cos \theta$

where E=electric field strength

A=area

$\theta$ =Angle between Electric field and area vector

$E=\frac{\phi }{A\cos (0)}$

$E=\frac{3\times 10^{-5}}{25\times 10^{-4}\times \cos(0)}$

$E=0.012\ N/C$

and Electric field by a uniformly charged sheet is given by

$E=\frac{\sigma }{2\epsilon_0}$

where $\sigma$ =charge density

$=\frac{\sigma }{\epsilon_0}$

$\sigma =0.012\times 8.85\times 10^{-12}$

$\sigma =2.12\times 10^{-13}\ C/m^2$

5 0

3 years ago

Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.

sveta [45]

Answer:

the two ice skater have the same momentum but the are in different directions.

Paula will have a greater speed than Ricardo after the push-off.

Explanation:

Given that:

Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.

A. Which skater, if either, has the greater momentum after the push-off? Explain.

The law of conservation of can be applied here in order to determine the skater that possess a greater momentum after the push -off

The law of conservation of momentum states that the total momentum of two or more objects acting upon one another will not change, provided there are no external forces acting on them.

So if two objects in motion collide, their total momentum before the collision will be the same as the total momentum after the collision.

Momentum is the product of mass and velocity.

SO, from the information given:

Let represent the mass of Paula with $m_{Pa}$ and its initial velocity with $u_{Pa}$

Let represent the mass of Ricardo with $m_{Ri}$ and its initial velocity with $u_{Ri}$

At rest ;

their velocities will be zero, i.e

$u_{Pa}$ = $u_{Ri}$ = 0

The initial momentum for this process can be represented as :

$m_{Pa}$ $u_{Pa}$ + $m_{Ri}$ $u_{Ri}$ = 0

after push off from each other then their final velocity will be $v_{Pa}$ and $v_{Ri}$

The we can say their final momentum is:

$m_{Pa}$ $v_{Pa}$ + $m_{Ri}$ $v_{Ri}$ = 0

Using the law of conservation of momentum as states earlier.

Initial momentum = final momentum = 0

$m_{Pa}$ $u_{Pa}$ + $m_{Ri}$ $u_{Ri}$ = $m_{Pa}$ $v_{Pa}$ + $m_{Ri}$ $v_{Ri}$

Since the initial velocities are stating at rest then ; u = 0

$m_{Pa}$ (0) + $m_{Pa}$ (0) = $m_{Pa}$ $v_{Pa}$ + $m_{Ri}$ $v_{Ri}$

$m_{Pa}$ $v_{Pa}$ + $m_{Ri}$ $v_{Ri}$ = 0

$m_{Pa}$ $v_{Pa}$ = - $m_{Ri}$ $v_{Ri}$

Hence, we can conclude that the two ice skater have the same momentum but the are in different directions.

B. Which skater, if either, has the greater speed after the push-off? Explain.

Given that Ricardo weighs more than Paula

So $m_{Ri} > m_{Pa}$ ;

Then $\mathsf{\dfrac{{m_{Ri}}}{m_{Pa} }= 1}$

The magnitude of their momentum which is a product of mass and velocity can now be expressed as:

$m_{Pa}$ $v_{Pa}$ = $m_{Ri}$ $v_{Ri}$

The ratio is

$\dfrac{v_{Pa}}{v_{Ri}} =\dfrac{m_{Ri}}{m_{Pa}} = 1$

$v_{Pa} >v_{Ri}$

Therefore, Paula will have a greater speed than Ricardo after the push-off.

6 0

3 years ago

Which of the following statements is true? (A) The measured value for the pressure of a gas in a container is almost independent

Lana71 [14]

Answer:

The correct answer to the question is

Both A and B are true

Explanation:

The particles of a gas are free to move to occupy the entire volume in which they are placed due to the smallerinter molecular forces holding them together hence due to the face that pressure is a measure of the Force per unit area that is Pressure P = ( Force F)/ (Area A) then the force per unit area, exerted on the all of the container by the gaseous particles which are colliding with each other and with the walss of the container is fairly constant through out the surface oof the container

In the case of the liquid which are held on together by more stronger forces, the force per nit area exerted by the liquid particle is transmitted from one particle to the next until it reaches the container's surface. Then remembering that the force of gravity on the liquid is acting in one direction (that is downwards) the sum of the fprce due to the weight incrreases as we progress deaper into the liquid hence the pressure increases per unit depth

5 0

3 years ago