Answer:![19.80\times 10^{10} J](https://tex.z-dn.net/?f=19.80%5Ctimes%2010%5E%7B10%7D%20J)
Explanation:
Given
mass of satellite ![m=7.45 \times 10^3 kg](https://tex.z-dn.net/?f=m%3D7.45%20%5Ctimes%2010%5E3%20kg)
orbital radius ![R=7.50\times 10^6 m](https://tex.z-dn.net/?f=R%3D7.50%5Ctimes%2010%5E6%20m)
mass of Earth ![M=5.97\times 10^{24} kg](https://tex.z-dn.net/?f=M%3D5.97%5Ctimes%2010%5E%7B24%7D%20kg)
Minimum amount of Energy to move Satellite from its orbit to an infinite distance is sum of Potential Energy + Kinetic Energy of Satellite
![W=U+K.E.](https://tex.z-dn.net/?f=W%3DU%2BK.E.)
![U=-G\frac{Mm}{R}](https://tex.z-dn.net/?f=U%3D-G%5Cfrac%7BMm%7D%7BR%7D)
![U=-6.67\times 10^{-11}\times \frac{5.97\times 10^{24}\times 7.45 \times 10^3}{7.50\times 10^6}](https://tex.z-dn.net/?f=U%3D-6.67%5Ctimes%2010%5E%7B-11%7D%5Ctimes%20%5Cfrac%7B5.97%5Ctimes%2010%5E%7B24%7D%5Ctimes%207.45%20%5Ctimes%2010%5E3%7D%7B7.50%5Ctimes%2010%5E6%7D)
![U=-\frac{296.658\times 10^{16}}{7.5\times 10^6}](https://tex.z-dn.net/?f=U%3D-%5Cfrac%7B296.658%5Ctimes%2010%5E%7B16%7D%7D%7B7.5%5Ctimes%2010%5E6%7D)
![U=-39.55\times 10^{10} J](https://tex.z-dn.net/?f=U%3D-39.55%5Ctimes%2010%5E%7B10%7D%20J)
![K.E.=\frac{1}{2}\times mv^2](https://tex.z-dn.net/?f=K.E.%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20mv%5E2)
Where ![v=orbital\ velocity](https://tex.z-dn.net/?f=v%3Dorbital%5C%20velocity%20)
![v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{6.67\times 10^{-11}\times 5.97\times 10^{24}}{7.50\times 10^6}}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cfrac%7BGM%7D%7Br%7D%7D%3D%5Csqrt%7B%5Cfrac%7B6.67%5Ctimes%2010%5E%7B-11%7D%5Ctimes%205.97%5Ctimes%2010%5E%7B24%7D%7D%7B7.50%5Ctimes%2010%5E6%7D%7D)
![v^2=5.30\times 10^7 m/s](https://tex.z-dn.net/?f=v%5E2%3D5.30%5Ctimes%2010%5E7%20m%2Fs)
![K.E.=\frac{1}{2}\times mv^2](https://tex.z-dn.net/?f=K.E.%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20mv%5E2)
![K.E.=\frac{1}{2}\times 7.45\times 10^3\times (5.30\times 10^7)=19.74\times 10^{10} J](https://tex.z-dn.net/?f=K.E.%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%207.45%5Ctimes%2010%5E3%5Ctimes%20%285.30%5Ctimes%2010%5E7%29%3D19.74%5Ctimes%2010%5E%7B10%7D%20J)
![W=-39.55\times 10^{10}+19.74\times 10^{10}](https://tex.z-dn.net/?f=W%3D-39.55%5Ctimes%2010%5E%7B10%7D%2B19.74%5Ctimes%2010%5E%7B10%7D)
![W=-19.80\times 10^{10} J](https://tex.z-dn.net/?f=W%3D-19.80%5Ctimes%2010%5E%7B10%7D%20J)
i.e.
Energy is required to provide to move Satellite out of its orbit
Answer:
Newton's Second Law of Motion says that acceleration (gaining speed) happens when a force acts on a mass (object). Riding your bicycle is a good example of this law of motion at work. Your bicycle is the mass. Your leg muscles pushing pushing on the pedals of your bicycle is the force.Newton's Second Law also says that the greater the mass of the object being accelerated, the greater the amount of force needed to accelerate the object. Say you have two identical bicycles that each have a basket. One bicycle has an empty basket. One bicycle has a basket full of bricks. If you try to ride each bicycle and you push on the pedals with the exact same strength, you will be able to accelerate the bike with the empty basket MORE than the bike with the basket full of bricks. The bricks add mass to the second bicycle. With bricks in the basket, you would have to apply more force to the pedals to make the bicycle with bricks in the basket move.
Explanation:
Explanation:
Initial speed of the rocket, u = 0
Acceleration of the rocket, ![a=25.4\ m/s^2](https://tex.z-dn.net/?f=a%3D25.4%5C%20m%2Fs%5E2)
Time taken, t = 3.39 s
Let v is the final velocity of the rocket when it runs out of fuels. Using the equation of kinematics as :
![v=u+at](https://tex.z-dn.net/?f=v%3Du%2Bat)
Let x is the initial position of the rocket. Using third equation of kinematics as :
![v^2=u^2+2ax_o](https://tex.z-dn.net/?f=v%5E2%3Du%5E2%2B2ax_o)
![x_o=\dfrac{v^2}{2a}](https://tex.z-dn.net/?f=x_o%3D%5Cdfrac%7Bv%5E2%7D%7B2a%7D)
Let
is the position at the maximum height. Again using equation of motion as :
![v^2-u^2=2a(x-x_o)](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2a%28x-x_o%29)
Now
and v and u will interchange
![u^2=2g(x-x_o)](https://tex.z-dn.net/?f=u%5E2%3D2g%28x-x_o%29)
![x=x_o+\dfrac{u^2}{2g}](https://tex.z-dn.net/?f=x%3Dx_o%2B%5Cdfrac%7Bu%5E2%7D%7B2g%7D)
![x=145.92+\dfrac{(86.10)^2}{2\times 9.8}](https://tex.z-dn.net/?f=x%3D145.92%2B%5Cdfrac%7B%2886.10%29%5E2%7D%7B2%5Ctimes%209.8%7D)
x = 524.14 meters
Hence, this is the required solution.
Answer:
1) HIGH IMPULSE, SHORT INSTANT
2) SMALL IMPULSE, MORE INSTANT OF TIME
more
Explanation:
Let's propose the solution of this exercise, let's use the relationship between momentum and momentum
I = Δp
F t = m v_f - m v₀
indicates that the carriage stops v_f = 0
F t = - m v₀
Therefore, if the impact time decreases, the force must increase and therefore the damages also
Car A collides with a concrete wall that is rigid, therefore the collision occurs in a very short time, car B collides with area, so the collision time is extended, now we can complete the sentences
1) Stunt Car A experiences a _HIGH IMPULSE_ over a SHORT INSTANT__ of time.
2)Stunt Car B experiences a _SMALL IMPULSE_____ over a _MORE INSTANT__ of time.
3) Because of the force experienced by Stunt Car A, it will sustain __MORE____ damage than Stunt Car B.a
3) more