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Luden [163]
3 years ago
9

(b) If the object is at 330 feet and its instantaneous velocity is 3 feet per minute at 30 minutes, what is the approximate posi

tion of the object at 32 minutes
Physics
1 answer:
ELEN [110]3 years ago
8 0

Answer:

The final position is 36 feet.

Explanation:

initial position, d = 330 feet

speed, v = 3 feet per minute

time, t = 30 minute

now the time is 32 minute

time interval = 2 minute

So, the distance in 2 minutes is

d' = 2 x 3 = 6 feet

So, the final position is

D = 30 + 6 = 36 feet

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A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .
maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

F_g = F_c

\frac{GmM}{r^2} = \frac{mv^2}{r}

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

\frac{GM}{r^2} = \frac{v^2}{r}

\frac{GM}{r} = v^2

v = \sqrt{\frac{GM}{r}}

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}

v = 4564.42m/s

From the speed it is possible to use find the formula, so

T = \frac{2\pi r}{v}

T = \frac{2\pi (6.371*10^6)}{4564.42}

T = 8770.05s\approx 146min\approx 2.4hour

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

KE = \frac{1}{2} mv^2

KE = \frac{1}{2} (100)(4564.42)^2

KE = 1.0416*10^9J

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

8 0
3 years ago
Please provide the steps
elena-s [515]

Answer:

wA - T =mA(a)

second option

Explanation:

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8 0
3 years ago
Which process is represented by the PV diagram shown below?
MrMuchimi

Answer: B. the isovolumetric process

Explanation:

In the graph given, the volume is constant throughout. It represents a constant volume process. Such processes are called the isovolumetric process or isochoric process.

<em>Hence, option B is the correct answer.</em>

Option A is incorrect because in an isobaric process, the pressure is constant.

Option C is incorrect because in an isothermal process, the temperature is constant.

Option D is incorrect because in an adiabatic process there is no heat transfer.

8 0
3 years ago
You find yourself in a place that is unimaginably hot and dense. A rapidly changing gravitational field randomly warps space and
Marta_Voda [28]

You find yourself in a place that is unimaginably <u>hot and dense</u>. A r<u>apidly changing</u><u> gravitational field</u><u> </u>randomly warps space and time. Gripped by these huge fluctuations, you notice that there is but a single, unified force governing the universe, you are in the early universe before the Planck time.

<h3>What is Planck time?</h3>

The Planck time is approximately<u> 10^-44 seconds</u>. The smallest time interval, or "zeptosecond," that has so far been measured is <u>10^-21 seconds</u>. A photon traveling at the speed of light would need one Planck time <u>to traverse a distance of one </u><u>Planck length</u>.

<h3>What is Planck length?</h3>

Planck units are a set of measuring units used only in particle physics and physical cosmology. They are defined in terms of <u>four universal </u><u>physical constants</u> in such a way that when expressed in terms of these units, these physical constants have the numerical value 1. These units are a system of natural units because its definition is <u>based on characteristics of nature</u>, more especially the characteristics of free space, rather than a selection of prototype object, as was the case with Max Planck's original 1899 proposal. They are pertinent to the study of unifying theories like quantum gravity.

To learn more about Plank time:

brainly.com/question/23791066

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8 0
2 years ago
In one cycle, a heat engine takes in 1000 J of heat from a high-temperature reservoir, releases 600 J of heat to a lower-tempera
Talja [164]

Answer:

η = 40 %  

Explanation:

Given that

Qa ,Heat addition= 1000  J

Qr,Heat rejection= 600 J

Work done ,W= 400 J

We know that ,efficiency of a engine given as

\eta=\dfrac{W(net)}{Q(heat\ addition)}

Now by putting the values in the above equation ,then we get

\eta=\dfrac{400}{1000}

η = 0.4

The efficiency in percentage is given as

η = 0.4  x 100 %

η = 40 %

Therefore the answer will be 40%.

4 0
3 years ago
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