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Vikentia [17]
3 years ago
12

A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite

from the planet is 6600 N. What is the kinetic energy of the satellite
Physics
1 answer:
sladkih [1.3K]3 years ago
7 0

Answer:

The  kinetic energy is KE  =  7.59  *10^{10} \  J

Explanation:

From the question we are told that

       The  radius of the orbit is  r =  2.3 *10^{4} \ km  = 2.3  *10^{7} \ m

       The gravitational force is  F_g  = 6600 \ N

The kinetic energy of the satellite is mathematically represented as

       KE  =  \frac{1}{2} * mv^2

where v is the speed of the satellite which is mathematically represented as

     v  = \sqrt{\frac{G  M}{r^2} }

=>  v^2  =  \frac{GM }{r}

substituting this into the equation

      KE  =  \frac{ 1}{2} *\frac{GMm}{r}

Now the gravitational force of the planet is mathematically represented as

      F_g  = \frac{GMm}{r^2}

Where M is the mass of the planet and  m is the mass of the satellite

 Now looking at the formula for KE we see that we can represent it as

     KE  =  \frac{ 1}{2} *[\frac{GMm}{r^2}] * r

=>    KE  =  \frac{ 1}{2} *F_g * r

substituting values

       KE  =  \frac{ 1}{2} *6600 * 2.3*10^{7}

         KE  =  7.59  *10^{10} \  J

 

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Alenkasestr [34]

Answer: 1.289 m

Explanation:

The path the cobra's venom follows since it is spitted until it hits the ground, is described by a parabola. Hence, the equations for parabolic motion (which has two components) can be applied to solve this problem:

<u>x-component: </u>

x=V_{o}cos\theta t  (1)

Where:

x is the horizontal distance traveled by the venom

V_{o}=3.10 m/s is the venom's initial speed

\theta=47\° is the angle

t is the time since the venom is spitted until it hits the ground

<u>y-component: </u>

y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}   (2)

Where:

y_{o}=0.44 m  is the initial height of the venom

y=0  is the final height of the venom (when it finally hits the ground)

g=-9.8m/s^{2}  is the acceleration due gravity

Let's begin with (2) to find the time it takes the complete path:

0=0.44 m+3.10 m/s sin\theta(47\°)+\frac{-9.8m/s^{2} t^{2}}{2}   (3)

Rewritting (3):

-4.9 m/s^{2} t^{2} + 2.267 m/s t + 0.44 m=0   (4)

This is a quadratic equation (also called equation of the second degree) of the form at^{2}+bt+c=0, which can be solved with the following formula:

t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a} (5)

Where:

a=-4.9 m/s^{2

b=2.267 m/s

c=0.44 m

Substituting the known values:

t=\frac{-2.267 \pm \sqrt{2.267^{2}-4(-4.9)(0.44)}}{2(-4.9)} (6)

Solving (6) we find the positive result is:

t=0.609 s (7)

Substituting (7) in (1):

x=(3.10 m/s)cos(47\°)(0.609 s)  (8)

We finally find the horizontal distance traveled by the venom:

x=1.289 m  

7 0
3 years ago
An automobile steering wheel is shown. What is the ideal mechanical advantage? If the AMA is 8, what is the efficiency of the st
Mars2501 [29]

1. Ideal Mechanical Advantage (IMA): 9

Explanation:

For a wheel and axle system like the steering wheel, the IMA is given by:

IMA=\frac{r_w}{r_a}

where

r_w is the radius of the wheel

r_a is the radius of the axle

For the steering wheel of the problem, we see that r_w = 18 cm and r_a=2 cm, so the IMA is

IMA=\frac{18 cm}{2 cm}=9


2. Efficiency: 88.9%

Explanation:

The efficiency of a system is defined as the ratio between the AMA (actual mechanical advantage) and the IMA:

\eta=\frac{AMA}{IMA}\cdot 100

In this problem, AMA=8 and IMA=9, so the efficiency is

\eta=\frac{8}{9}\cdot 100=88.9\%


6 0
3 years ago
When there's a hazard ahead, it's almost always quicker for you to _________ than to come to a full stop.
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When there's a hazard ahead, it's almost always quicker for you to steer away than to come to a full stop.

<h3>What is an hazard?</h3>

Hazard refers to any obstacle or other feature which causes risk or danger.

Living organisms respond to hazards via the production of adrenaline hormone. This hormone causes a flight response away from the hazard.

Therefore, when there's a hazard ahead, it's almost always quicker for you to steer away than to come to a full stop.

Learn more about hazards at: brainly.com/question/5338299

5 0
1 year ago
A bowling ball of mass 5 kg rolls down a slick ramp 20 meters long at a 30 degree angle to the horizontal. What is the work done
Elena-2011 [213]

Answer:

The work done by gravity during the roll is 490.6 J

Explanation:

The work (W) is:

W = F*d

<em>Where</em>:

F: is the force

d: is the displacement = 20 m

The force is equal to the weight (W) in the x component:

F = W_{x} = mgsin(\theta)

<em>Where:</em>

m: is the mass of the bowling ball = 5 kg

g: is the gravity = 9.81 m/s²    

θ: is the degree angle to the horizontal = 30°        

F = mgsin(\theta) = 5 kg*9.81 m/s^{2}*sin(30) = 24.53 N    

Now, we can find the work:

W = F*d = 24.53 N*20 m = 490.6 J      

Therefore, the work done by gravity during the roll is 490.6 J.

I hope it helps you!

6 0
3 years ago
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solmaris [256]

Answer:

There are seven principles that form the content grounds of our teaching framework:

Non-maleficence. ...

Beneficence. ...

Health maximisation. ...

Efficiency. ...

Respect for autonomy. ...

Justice. ...

Proportionality.

5 0
2 years ago
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