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Oksi-84 [34.3K]
3 years ago
13

The density of atmosphere on a certain planet is found to decrease as altitude increases what is the type of relationship betwee

n the height and the atmospheric density and which segment can give the most accurate approximations for the inter lipo later values
Physics
1 answer:
8_murik_8 [283]3 years ago
6 0
? Is there options ??????
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An object travels along a straight, horizontal surface with an initial speed of 2 ms. The position of the object as a function o
pashok25 [27]

Answer:

The options are not provided, so i will answer in a general way.

We know that:

The movement is along a straight horizontal surface, then we have one-dimensional motion.

The speed is 2m/s

We want a graph of position vs time.

Now, remember the relation:

Distance = Speed*Time

Then we can write the position as a function of time as:

P(t) = 2m/s*t + P0

Where t is our variable, that represents time in seconds, and P0 is the position at time t = 0seconds, we can assume that this is zero.

Then the equation is:

P(t) = 2m/s*t

And the graph is something like:

3 0
3 years ago
How can we protect space shuttles or astronauts from space radiation in the absence of the atmospheric layer?
antiseptic1488 [7]

Answer:

1. In general, the best shields will be able to block a spectrum of radiation. Aboard the space station, the use of hydrogen-rich shielding such as polyethylene in the most frequently occupied locations, such as the sleeping quarters and the galley, has reduced the crew's exposure to space radiation.

2. It absorbs harmful ultraviolet radiation from the sun, helps keep Earth's surface warm via the greenhouse effect, and reduces temperature extremes between day and night. ... So, thanks to gravity, although some of Earth's atmosphere is escaping to space, most is staying here.

hope it helps. please mark me as brainliest and follow me ❤️

3 0
2 years ago
Which branch of science involves the study of matter and energy
Olin [163]
Physics<span> is a natural science that involves the study of matter and its motion through space time, along with related concepts such as energy and force.</span>
6 0
3 years ago
Read 2 more answers
a stone is vertically thrown upward with the velocity of 72km/hr find the maximum height reached the height​
dolphi86 [110]

Answer:

40m

Explanation:

v^2=gh

h=v^2/g

h=20^2/10

h=400/10

h=40

3 0
2 years ago
Derive the equation of motion of the block of mass m1 in terms of its displacement x. The friction between the block and the sur
Alenkasestr [34]

Answer:

the equivalent mass : m_e = m_1+m_2+\frac{I}{R^2}

the equation of the motion of the block of mass m_1 in terms of its displacement is = (m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

Explanation:

Let use m₁ to represent the mass of the block and m₂ to represent the mass of the cylinder

The radius of the cylinder  be = R

The distance between the center of the pulley to center of the block to be = x

Also, the angles of inclinations of the cylinder and the block with respect to the ground to be \phi and \beta respectively.

The velocity of the block to be = v

The equivalent mass of the system = m_e

In the terms of the equivalent mass, the kinetic energy of the system can be written as:

K.E = \frac{1}{2} m_ev^2       --------------- equation (1)

The angular velocity of the cylinder = \omega  :  &

The inertia of the cylinder about its center to be = I

The angular velocity of the cylinder can be written as:

v = \omega R

\omega =\frac{v}{R}

The kinetic energy of the system in terms of individual mass can be written as:

K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I\omega^2

By replacing \omega with \frac{v}{R} ; we have:

K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I(\frac{v}{R})^2

K.E = \frac{1}{2}(m_1+ m_2+ \frac{I}{R} )v^2   ------------------ equation (2)

Equating both equation (1) and (2); we have:

m_e = m_1+m_2+\frac{I}{R^2}

Therefore, the equivalent mass : m_e = m_1+m_2+\frac{I}{R^2}    which is read as;

The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by  the square of the radius.

The expression for the force acting on equivalent mass due to the block is as follows:

f_{block }=m_1gsin \beta

Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:

f_{cylinder} = m_2gsin \phi

Equating the above both equations; we have the equation of motion of the  equivalent system to be

m_e \bar x = f_{cylinder}-f_{block}

which can be written as follows from the previous derivations

(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

Finally; the equation of the motion of the block of mass m_1 in terms of its displacement is = (m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

8 0
3 years ago
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