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Lesechka [4]
2 years ago
7

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

Physics
1 answer:
maria [59]2 years ago
8 0

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.

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KengaRu [80]

If the mass of the object and the volume of the object is determined;

Then, the density of the object is determined by taking the ratio of the mass and volume.

<h3>What is density of an object?</h3>

The density of an object is the ratio of the mass and volume of that object.

Mathematically;

  • Density = mass/volume

To determine the density of an object therefore, the physical characteristics of mass and the volume of the object are measured.

The mass of the object is obtained using a scale or a balance.

The volume of the object if a solid is obtained using a displacement bottle. If it is a liquid, a measuring cylinder is used.

The density of the object is then obtained by taking the ratio of the mass and the volume of the object.

In conclusion, the density of an object is determined from the volume and mass ratio.

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3 0
1 year ago
If a cliff jumper leaps off the edge of a 100m cliff, how long does she fall before hitting the water? (assume zero air resistan
andrew-mc [135]
<h2>Answer:</h2>

<em>Hello, </em>

<h3><u>QUESTION)</u></h3>

Assuming that the initial velocity of the jumper is zero, on Earth any freely falling object has an acceleration of 9.8 m/s².  

<em>✔ We have : a = v/Δt = ⇔ Δt = v/a </em>

  • Δt = (√2xgxh)/9,8
  • Δt = (14√10)/9,8
  • Δt ≈ 4,5 s

4 0
2 years ago
The purpose of this lab is to explore the various ways to calculate projectile velocity using horizontal, vertical and angle inf
kolezko [41]

Answer: A projectile is any object in which the only force is gravity

Explanation: Equations on how to calculate projectile velocity is stated below:

The initial velocity Vo being a vector quantity, has two componentsVox and Voy  

V0x = V0 cos(θ) 

V0y = V0 sin(θ) 

The acceleration A is a also a vector with two components Axand Ay given

Ax = 0 and Ay = - g = - 9.8 m/s2 

Along the x axis the acceleration is equal to 0 and therefore the velocity Vx is constant  

Vx = Vocos(θ) 

Along the y axis, the acceleration is uniform and equal to - g and the velocity at time t is g

Vy = Vo sin(θ) - g t 

Along the x axis the velocity Vx is constant and therefore the component x of the displacement is

x = Vocos(θ) t 

Along the y axis, the motion is of uniform acceleration and the y component of the displacement is

y = Vo sin(θ) t - (1/2) g t2 

3 0
3 years ago
How much water will flow in 30 secs through 200 mm of capillary tube of 1.50 mm in diameter, if the pressure difference across t
Paladinen [302]

The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

Qo=1.6 \times 10^{2} \mathrm{~mL}

<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>

\begin{aligned}\Delta P &=6660 \mathrm{~m} / \mathrm{m}^{2} \\\mu &=8.01 \times 10^{-4} \text { Pas } \\t &=30 \mathrm{~s} \\L &=200 \mathrm{~mm}=200 \times 10^{-3} \mathrm{~m} \\D &=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \Rightarrow \gamma=\frac{1.5 \times 10^{-3}}{2} \mathrm{~m}\end{aligned}

Generally, the equation for Rate of flow of Liquid is  mathematically given as

\\$$Q=\frac{\pi r^{4} \times \Delta P}{8 \mu L}

$$

Where dP is pressure difference r is the radius

\mu is the viscosity of water

L is the length of the pipe

Q=\frac{\pi \times\left(\frac{1.5 \times 10^{-3}}{2}\right)^{4} \times 6660}{8 \times 8.01 \times 10^{-4} \times 200 \times 10^{-3}}

Q=5.2 \mathrm{~mL} / \mathrm{s}

In $30s the quantity that flows out of the tube

&Qo=5.2 \times 30 \\&Qo=1.6 \times 10^{2} \mathrm{~mL}

In conclusion, the quantity that flows out of the tube

Qo=1.6 \times 10^{2} \mathrm{~mL}

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5 0
1 year ago
It takes you 8.3 min to walk with an average velocity of 1.6 m/s to the north from the bus stop to the museum entrance. How far
Anarel [89]

solution:

1.6 m/s = 96 m/min (in other words, 1.6 m/s x 60 s/min)  

96 m/min x 8.3 min = 796.8 m

s=ut +\frac{1}{2}at^2\\there is no accleration mentioned so,\\s= uv\\8.3\times60=498(s)\\510\times1.6=816(m)


3 0
3 years ago
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