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ValentinkaMS [17]
1 year ago
8

Q C Many people assume air resistance acting on a moving object will always make the object slow down. It can, however, actually

be responsible for making the object speed up. Consider a 100 - kg Earth satellite in a circular orbit at an altitude of 200 km . A small force of air resistance makes the satellite drop into a circular orbit with an altitude of 100km. (e) Show that the system has lost mechanical energy and find the amount of the loss due to friction.
Physics
1 answer:
OLga [1]1 year ago
7 0

$E_i-E_f=-17.94 \times 10^9 \mathrm{~J}-\left(-18.30 \times 10^9 \mathrm{~J}\right)=3.60 \times 10^8 \mathrm{~J}$.

The most effective forces on the object are the backward force of air resistance relatively very small in magnitude, and the force of gravity. Because the spiral path of the satellite is not perpendicular to the gravitational force, one element of the gravitational force pulls forward. at the satellite to do fantastic work & make its speed increase.

<h3>What is called gravitational force?</h3>

Gravity, additionally referred to as gravitation, is a force that exists amongst all material gadgets withinside the universe. For any objects or particles having nonzero mass, the force of gravity tends to draw them in the direction of each other. Gravity operates on objects of all sizes, from subatomic particles to clusters of galaxies.

To learn more about gravitational force, visit;

brainly.com/question/9266911

#SPJ4

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I NEED HELP PLEASE, THANKS! :)
Zina [86]

Answer:

charge C = greatest net force

charge B = the smallest net force

ratio  = 9 : 1

Explanation:

we know that in Electrostatic Forces, when 2 charges are at same sign then they repel each other and if they are different signed charges then they attract each other

so as per Coulomb's formula of Electrostatic Forces

F = \frac{k\ q_1\ q_2}{r^2}     .....................1

and here k is 9 × 10^9 N.m²/c² and we consider each charge at distance d

so two charge force at A to B is

F1 = \frac{k\ q^2}{d^2}

and force between charges at A to C, at 2d distance

F1 = \frac{k\ q^2}{(2d)^2}  =  \frac{k\ q^2}{4d^2}

force between charges at A to D,  3d distance

F1 = \frac{k\ q^2}{(3d)^2}  = \frac{k\ q^2}{9d^2}  

so

Charge a It receives force to the left from b and c and to the right from d

so at a will be

F(a)  = -F1 - F2 + F3             ....................2

put here value

F(a) = -\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}

solve it

F(a) = \frac{k\ q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})  

F(a) = -\frac{41}{36}\ F1   = 1.13 F1  

and

Charge b It  receives force to the right from a and d and to the left from c

F(b) = F1 - F1 + F2            ....................3

F(b)  =  \frac{k\ q^2}{d^2}-\frac{k\ q^2}{d^2}+\frac{k\ q^2}{4d^2}    

F(b)  = \frac{1}{4} \ F1    =  0.25 F1

and

Charge c It receives forces to the right from all charges.

F(c) = F2 + F 1 + F 1      ....................4

F(c) = \frac{k\ q^2}{4d^2}+\frac{k\ q^2}{d^2}+\frac{k\ q^2}{d^2}      

F(c) =  \frac{9}{4} \ F1   = 2.25 F1

and

Charge d It receives forces to the left from all charges

F(d) = - F3 - F2 -F 1      ....................5

F(d) = -\frac{k\ q^2}{9d^2}-\frac{k\ q^2}{4d^2}-\frac{k\ q^2}{d^2}  

so

F(d) = -\frac{49}{36} \ F1    = 1.36 F1

and

now we get here ratio of the greatest to the smallest net force that is

ratio = \frac{2.25}{0.25}

 ratio  = 9 : 1

5 0
3 years ago
What is the momentum of an object with 5.22 kg and 1.7 m/s
BlackZzzverrR [31]

Answer:

8.874

Explanation:

You need to times 5.22 kg and 1.7 m/s to get 8.874.

8 0
3 years ago
The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.32 with the floor. If t
coldgirl [10]

Answer:

The shortest braking distance is 35.8 m

Explanation:

To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down

On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis

    Y axis

     N- W = 0

     N = W = mg

  X axis

     -Fr = m a

     -μ N = m a

     -μ mg = ma

     a = μ g

     a  = - 0.32 9.8

     a =  - 3.14 m/s²

We calculate the distance using the kinematics equations

    Vf² = Vo² + 2 a x

     x = (Vf² - Vo²) / 2 a

When the train stops the speed is zero (Vf = 0)

 Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s

     x = ( 0 - 15²) / 2 (-3.14)

     x=  35.8 m

The shortest braking distance is  35.8 m

7 0
3 years ago
Help plzzzzzzz i need thissssssssss
frutty [35]

Answer:

The final graph

Explanation:

The graph that curves downwards is negative acceleration. While the position decreases the slop increases.

8 0
3 years ago
Is it possible for a tennis ball to have more momentum than a cannonball? explain ​
Olenka [21]
Well you know how it
8 0
3 years ago
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