The small spherical planet called "Glob" has a mass of 7.88×10^18 kg and a radius of 6.32×10^4 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×10^3 m, above the surface of the planet, before it falls back down.

1. What was the initial speed of the rock as it left the astronaut's hand? (Glob has no atmosphere, so no energy is lost to air friction. G = 6.67×10^-11 Nm2/kg2.)

2. A 36.0 kg satellite is in a circular orbit with a radius of 1.45×10^5 m around the planet Glob. Calculate the speed of the satellite.

Answers

Answer 1

Answer: The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.

1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.

2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.

Explanation: To find the answer, we need to know about the different equations of planetary motion.

How to find the initial speed of the rock as it left the astronaut's hand?We have the expression for the initial velocity as,

                           [tex]v=\sqrt{2gh}[/tex]

Thus, to find v, we have to find the acceleration due to gravity of glob. For this, we have,

                       [tex]g_g=\frac{GM}{r^2} =\frac{6.67*10^{-11}*7.88*10^{18}}{(6.32*10^4)^2}= 0.132[/tex]

Now, the velocity will become,

                        [tex]v=\sqrt{2*0.132*1.44*10^3} =19.46 m/s[/tex]

How to find the speed of the satellite?As we know that, by equating both centripetal force and the gravitational force, we get the equation of speed of a satellite as,

                       [tex]v=\sqrt{\frac{GM}{r} } =\sqrt{\frac{6.67*10^{-11}*7.88*10^{18}}{1.45*10^5} } =3.624km/s[/tex]

Thus, we can conclude that,

1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.

2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.

Learn more about the equations of planetary motion here:

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Answer 2

The tiny planet known as "Glob" has a radius of 6.32× 10^4 meters and a mass of 7.88× 10^18 kg. On Glob's surface, an astronaut launches a rock straight upward. Before falling back down, the rock rises to a maximum height of 1.44×10^3 m above the planet's surface.

1) The rock was moving at 19.46 m/s when it first left the astronaut's palm.

2) A 36.0 kg spacecraft is orbiting the planet Glob in a sphere with a radius of 1.45 105 meters. The satellite is moving at 3.624 km/s at that point.

Understanding the planetary motion equations is necessary in order to determine the solution.

How to determine the rock's original speed when it left the astronaut's hand?The starting velocity's expression is as follows:

                                [tex]V=\sqrt{2gh}[/tex]

So, in order to determine v, we must determine the acceleration of glob caused by gravity. We already have,

                     [tex]a=\frac{GM}{r^2} =\frac{6.67*10^{-11}*7.88*10^{18}}{(6.32*10^4)^2} \\a=0.132m/s^2[/tex]

The velocity will now change to,

                   [tex]V=\sqrt{2*0.132*1.44*10^3} =19.46m/s[/tex]

How can I determine the satellite's speed?As we are aware, the centripetal force and gravitational force are equivalent, and thus leads to the following satellite speed equation:

                         [tex]v=\sqrt{\frac{GM}{r} } =3,624km/s\\where,\\M=7.88*10^{18}kg[/tex]

Consequently, we can say that

1) The rock was moving at 19.46 m/s when it first left the astronaut's palm.

2) A 36.0 kg spacecraft is orbiting the planet Glob in a sphere with a radius of 1.45 105 meters. The satellite is moving at 3.624 km/s at that point.

Learn more about the planetary motion here:

https://brainly.com/question/28108487

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Related Questions

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What can you infer about the strength and direction of forces experienced by the pod and space station when they collided? What evidence from today’s activities supports your inference?

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Answer:

In the collision, the strength of the force exerted on the pod is greater than the strength of the force exerted on the space station, but those forces are exerted in opposite directions.

Explanation:

a body of mass 20kg initially at rest is subjected to a force of 40N for 1sec calculate the change in kinetic energy showing the solution​

Answers

Answer:

Change in KE is 40 J

Explanation:

Recall that the impulse exerted on an object equal the change of momentum of the object (ΔP), which in time is defined as the product of the force exerted on it times the time the force was acting:

Change in momentum is:   ΔP = F * Δt

In our case,

ΔP = 40 N * 1 sec = 40 N s

Since the object was initially at rest, its initial momentum was zero, and the final momentum should then be 40 N s.

So, the initial KE was 0, and the final (KEf) can be calculated using:

KEf = 1 /(2 m) Pf^2 = 1 / (40) 40^2 = 40 J

So, the change in kinetic energy is:

KEf - KEi = 40 J - 0 j = 40 J

What kind of reasoning is most often used to form hypotheses?
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Deductive reasoning

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The mystery that Greene discusses is that dark energy is causing the expansion of the universe to accelerate. However, this cannot be explained by the laws of Physics.

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Thermodynamic Processes

Two moles of a monatomic ideal gas at (5 MPa, 5 L) is expanded isothermally until the volume is doubled (step 1). Then it is cooled isochorically until the pressure is 1 MPa (step 2). The temperature drops in this process. The gas is now compressed isothermally until its volume is back to 5 L, but its pressure is now 2 MPa (step 3). Finally, the gas is heated isochorically to return to the initial state (step 4). (a) Draw the four processes in the pV plane. (b) Find the total work done by the gas.

Answers

Answer:

A. Part a is the attachment

B. total work = 10.4kj

Explanation:

[tex]workdone=nRT1ln\frac{Vb}{Va}[/tex]

T1 = constant temperature

nRT1 = PaVa = PbVb

We write equation as

[tex]workdone =(PaVa)ln\frac{Vb}{Va}[/tex]

5ma = Pa, 5L = Va, Vb = 10L(temperature is doubled)

[tex]w1 = workdone =(5mpa*5L)ln\frac{10L}{5L}[/tex]

W1 = 25 ln2

W1 = 25 x 0.693

= 17.327kj

The isochoric expansion has no change in volume. So,

W2 = 0

Isothermal compression

[tex]w3=nRT3ln\frac{Vd}{Vc}[/tex]

T3 = constant temperature

nRT3 = PcVc = PdVd

[tex]workdone=(PcVc)ln\frac{Vd}{Vc}[/tex]

Pc = 1mpa Vc = 10L Vd = 5L

[tex]w3=(1)(10)ln\frac{5L}{10L}[/tex]

= 10x-0.693

= -6.93kj

Isochoric compression has no change in volume. Workdone w4 = 0

Total workdone = w1 + w2 + w3 + w4

= 17.33 + 0 + (-6.93) + 0

= 10.4kj

Galileo
o did not believe friction existed
o believed that friction stopped objects in motion
o believed that friction kept objects in motion
О
assumed that in a frictionless environment objects would never move

Answers

Answer:

object would move but it could be difficult to slow down or stop.

Formula One racers speed up much more quickly than normal passenger vehicles, and they also can stop in a much shorter distance. A Formula One racer traveling at 90m/s can stop in a distance of 110m. What is the magnitude of the car's acceleration as it slows during braking?

Answers

Answer:

The magnitude of the car's acceleration as it slows during braking is 36.81 m/s²

Explanation:

From the question, the given values are as follows:

Initial velocity, u = 90 m/s

final velocity, v = 0 m/s

distance, s = 110 m

acceleration, a = ?

Using the equation of motion, v² = u² + 2as

(90)² + 2 * 110 * a = 0

8100 + 220a = 0

220a = -8100

a = -8100/220

a = -36.81 m/s²

The value for acceleration is negative showing that car is decelerating to a stop. The magnitude of the car's acceleration as it slows during braking is therefore 36.81 m/s²

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Answer:

D, the acceleration  of A is twice that of b.

Explanation: in four seconds b got to ten, in two seconds a got to 20. Going  10m/s faster in half the time is going twice the acceleration

alex often draws his dream house​

Answers

Answer:

hopefully alex quackity hahhaa

Explanation:

i hope this was free points and not an actual thing

Answer:

cool, cool for alex .....

The force that is applied to other objects from the electric field *
A. ions
B. electric field
C. electric force
D. static charge

Answers

electric force hope this helps!!!

Two students on ice skates stand one behind the other. Student 2 pushes student 1 in the back; both students move away from each other. What law of motion is this. (Newton's laws)

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Answer:

forcing in act

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Calibration graphs can be used to determine unknown concentrations in electrochemical ​

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Matter is made of small particles to small to be seen. Which of these best describe evidence of this statement? 1. Tara’s crayons melted when she left them under the sun. 2. Kerris glass of water overflowed when she added ice. 3. Sams basketball expands as he pumps air into it. 4. Stephanie dropped a vase and it broke into pieces.

Answers

Answer:

Explanation:

I think the answer is statement no 3.

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Answer:

1 Tara's crayons melted when she left them under the sun

A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle are both traveling at the same speed of 19.0 m/s , and the distance between them is 52.0 m . After t1 = 3.00 s , the motorcycle starts to accelerate at a rate of 4.00 m/s^2. The motorcycle catches up with the car at some time t2.

Required:
a. How long does it take from the moment when the motorcycle starts to accelerate until it catches up with the car?
b. How far does the motorcycle travel from the moment it starts to accelerate (at time t1) until it catches up with the car (at time t2)?

Answers

Answer:

a) 5.09 seconds

b) 107.07 meters

Explanation:

a) As we know

[tex]t_2- t_1 = \sqrt{\frac{2 X}{a} }[/tex]

Substituting the given values we get

[tex]t_2 - t_1 = \sqrt{\frac{2 * 52}{4} } \\t_2 - t_1 = 5.09[/tex]

It takes 5 .09 s for the motorcycle to accelerate until it catches up with the car

b)

[tex]X_{t`2} = v_i \sqrt{\frac{2X}{a} } + 0.5 a\sqrt{\frac{2X}{a} }\\X_{t`2} = (v_i + 0.5 a) \sqrt{\frac{2X}{a} }\\X_{t`2} = ( 19 + 2) \sqrt{\frac{2* 52}{4} }\\X_{t`2} = 21 * 5.09\\X_{t`2} = 107.07[/tex]

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Answer:

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Canon launch is a 4.0 kg bowling ball with 50 J of kinetic energy what is the bowling ball speed

Answers

Answer:

5 m/s

Explanation:

50=1/2*4v^2

4*1/2=2

25*2=50

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What is the average speed of an Olympic sprinter that runs 100 m in 9.88 s?

Answers

Answer:

speed = 10.1215 m/s

Explanation:

speed = distance / time

speed = 100 / 9.88 = 10.1215 m/s

Let’s look at a radio-controlled model car. Suppose that at time t1=2.0st1=2.0s the car has components of velocity vx=1.0m/svx=1.0m/s and vy=3.0m/svy=3.0m/s and that at time t2=2.5st2=2.5s the components are vx=4.0m/svx=4.0m/s and vy=3.0m/svy=3.0m/s . Find (a) the components of average acceleration and (b) the magnitude and direction of the average acceleration during this interval.

Answers

Answer:

[tex]a_x=6\ \text{m/s}^2[/tex] and [tex]a_y=0\ \text{m/s}^2[/tex]

Magnitude of accleration is [tex]6\ \text{m/s}^2[/tex] and the direction is [tex]0^{\circ}[/tex]

Explanation:

[tex]t_1=2\ \text{s}[/tex]

[tex]v_x=1\ \text{m/s}[/tex]

[tex]v_y=3\ \text{m/s}[/tex]

[tex]t_2=2.5\ \text{s}[/tex]

[tex]v_x=4\ \text{m/s}[/tex]

[tex]v_y=3\ \text{m/s}[/tex]

Average acceleration in the different axes

[tex]a_x=\dfrac{\Delta v_x}{\Delta t}\\\Rightarrow a_x=\dfrac{4-1}{2.5-2}\\\Rightarrow a_x=6\ \text{m/s}^2[/tex]

[tex]a_y=\dfrac{\Delta v_y}{\Delta t}\\\Rightarrow a_y=\dfrac{3-3}{2.5-2}\\\Rightarrow a_y=0\ \text{m/s}^2[/tex]

The components of the acceleration is [tex]a_x=6\ \text{m/s}^2[/tex] and [tex]a_y=0\ \text{m/s}^2[/tex]

The magnitude of acceleration

[tex]a=\sqrt{a_x^2+a_y^2}\\\Rightarrow a=\sqrt{6^2+0^2}\\\Rightarrow a=6\ \text{m/s}^2[/tex]

Direction

[tex]\theta=\tan^{-1}\dfrac{a_y}{a_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{0}{6}\\\Rightarrow \theta=0^{\circ}[/tex]

The magnitude of accleration is [tex]6\ \text{m/s}^2[/tex] and the direction is [tex]0^{\circ}[/tex].

A proud new Jaguar owner drives her car at a speed of 35 m/s into a corner. The coefficients of friction between the road and the tires are 0.70 (static) and 0.40 (kinetic). What is the minimum radius of curvature for the corner in order for the car not to skid

Answers

Answer:

178.6 m

Explanation:

Since the car moves in a circular path, it experiences a centripetal force, F = mv²/r where m = mass of car, v = speed of car = 35 m/s and r = radius of curvature of path.

Now, for the car not to skid, this centripetal force must be equal to the frictional force, F' acting in the opposite direction.

So, F' = μN where μ = coefficient of static friction(since the car does not move in this direction) and N = normal force = mg where m = mass of car and g = acceleration due to gravity = 9.8m/s²

F' = μmg

Since F = F'

mv²/r = μmg

dividing both sides by m, we have

v²/r = μg

multiplying both sides by r, we have

v² = μgr

dividing both sides by μg, we have

r = v²/μg

Here we use μ = coefficient of static friction(since the car does not move in this direction) = 0.70. Substituting the other variables into the equation, we have

r = v²/μg

r = (35 m/s)²/(0.70 × 9.8m/s²)  

r = 1225 m²/s²/6.86m/s²)  

r = 178.6 m

So, the minimum radius of curvature of the corner is 178.6 m

Acceleration is the rate ot change of the velacity a -dejdt so it is the slope of the Velocity vs. Time graph Because it is dficult to drag the person in a consistent and reproducible way use the Expression Evakaator under the Special Features menu for this question lick Reset A and type in the hr on z t * t * t " t in the Expression Evaluator Click the Play button and let the simulation run roughly 5 sin ulation seconds before ressing the Pause but use the zoom buttons to a 쪄 the p s they the screen You should see 8 p at s ar l what you got in the previous question, but much smoother Look at the Postion vs Time. Velocity vs Time and Acceleration vs. Time piets h
a) the velocity is zero but the acceleration is negative
When the person is 8 to to the tight of the origin
b) the velocity is zero but the acceleration is positive
c) both the velocity and the acceleration are zero
d) both the velocity and the acceleraton are nonzero

Answers

Answer:

a) the body  is changing direction,

b)the body must go to the left and the acceleration to the right

c) the movement has not started.

d) all points of the motion

Explanation:

In this exercise you are asked to find in which position you have the following characteristics of the movement

a) The velocity is zero and the acceleration is negative

This is when the body reaches the end of the travel and turns around, in this case the speed is zero and the acceleration has the opposite direction to the movement.

In this case the body moves to the right and the acceleration is to the left, therefore the speed decreases

b) The velocity is zero, but the acceleration is positive

This occurs at the points where the speed is changing direction, specifically for this case the body must go to the left and the acceleration to the right

c) Both are zero

This only occurs where the body is stopped and the movement has not started.

d) both the velocity and the relation are nonzero.

This is at all points of the motion since the velocity is constantly changing as long as there is an acceleration

what is borh's postulates for the hydrogen atom​

Answers

Answer:

An atom has a number of stable orbits in which an electron can reside without the emission of radiant energy. ... Each orbit corresponds, to a certain energy level.

Explanation:

Hope it is helpful....

How does the Law of Conservation of Energy (or energy transformation) relate to the home?

Answers

Answer:

"The law of conservation of energy states that energy can neither be created nor destroyed - only converted from one form of energy to another. This means that a system always has the same amount of energy, unless it's added from the outside. ... The only way to use energy is to transform energy from one form to another."

Explanation:

Brainliest?

Which of the following is not an example of energy transfer?
A. Riding a bike down a hill.
B. Sliding a table across a floor.
C. Holding a sign in the air.

Answers

Answer:

c.holding a sign in the air

Explanation:

because b is kinetic energy and a is also kinetic energy

C holding a sign in the air

HURRY!!!
I need helppppp!!!

Answers

Answer:

ok i think the answer would be C. or B. hope im right

Three joules of work is needed to shift 10 C of charge from one place to another. The potential difference between the places is ______. *
0.3 V
0.5 V
5 V
3 V​

Answers

Answer:

The potential difference between the places is 0.3 V.

∴ 1st option i.e. 0.3V is the correct option.

Explanation:

Given

Work done W = 3J

Amount of Charge q = 10C

To determine

We need to determine the potential difference V between the places.

The potential difference between the two points can be determined using the formula

Potential Difference (V) = Work Done (W) / Amount of Charge (q)

or

[tex]\:V\:=\:\frac{W}{q}[/tex]

substituting W = 3 and q = 10 in the formula

[tex]V=\frac{3}{10}[/tex]

[tex]V=0.3[/tex] V

Therefore, the potential difference between the places is 0.3 V.

∴ 1st option i.e. 0.3V is the correct option.

Plzz help me with this
I’ll give brainliest

Answers

Answer:

B. Objects with more mass have more gravitational force acting upon them.

Answer:

Should be A but it can be B as well.

something that orbiys other things in space​

Answers

Answer: well we all orbit the sun all the planets do so the

SuN

Explanation: two words common sense

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