Based on the calculations, the speed required for this satellite to stay in orbit is equal to 1.8 × 10³ m/s.
<u>Given the following data:</u>
- Gravitational constant = 6.67 × 10⁻¹¹ m/kg²
- Mass of Moon = 7.36 × 10²² kg
- Distance, r = 4.2 × 10⁶ m.
<h3>How to determine the speed of this satellite?</h3>
In order to determine the speed of this satellite to stay in orbit, the centripetal force acting on it must be sufficient to change its direction.
This ultimately implies that, the centripetal force must be equal to the gravitational force as shown below:
Fc = Fg
mv²/r = GmM/r²
<u>Where:</u>
- m is the mass of the satellite.
Making v the subject of formula, we have;
v = √(GM/r)
Substituting the given parameters into the formula, we have;
v = √(6.67 × 10⁻¹¹ × 7.36 × 10²²/4.2 × 10⁶)
v = √(1,168,838.095)
v = 1,081.13 m/s.
Speed, v = 1.8 × 10³ m/s.
Read more on speed here: brainly.com/question/20162935
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Explanation:
We have,
The initial position of an object is zero.
The starting velocity is 3 m/s and the final velocity was 10 m/s.
The object moves with constant acceleration..
The area covered under the velocity-time graph gives displacement of the object. The correct option is "the area of the rectangle plus the area of the triangle under the line".
Answer:1.71 m/s
Explanation:
Given
mass of Susan ![m=12 kg](https://tex.z-dn.net/?f=m%3D12%20kg)
Inclination ![\theta =30^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%20%3D30%5E%7B%5Ccirc%7D)
Tension ![T=29 N](https://tex.z-dn.net/?f=T%3D29%20N)
coefficient of Friction ![\mu =0.18](https://tex.z-dn.net/?f=%5Cmu%20%3D0.18)
Resolving Forces Along x axis
![F_x=T\cos \theta -f_r](https://tex.z-dn.net/?f=F_x%3DT%5Ccos%20%5Ctheta%20-f_r)
where
![F_y=mg-N-T\sin \theta](https://tex.z-dn.net/?f=F_y%3Dmg-N-T%5Csin%20%5Ctheta%20)
since there is no movement in Y direction therefore
![N=mg-T\sin \theta](https://tex.z-dn.net/?f=N%3Dmg-T%5Csin%20%5Ctheta%20)
and ![f_r=\mu N](https://tex.z-dn.net/?f=f_r%3D%5Cmu%20N)
Thus ![F_x=T\cos \theta -\mu N](https://tex.z-dn.net/?f=F_x%3DT%5Ccos%20%5Ctheta%20-%5Cmu%20N)
![F_x=25.114-18.558](https://tex.z-dn.net/?f=F_x%3D25.114-18.558)
![F_x=6.556 N](https://tex.z-dn.net/?f=F_x%3D6.556%20N)
Work done by applied Force is equal to change to kinetic Energy
![F_x\cdot x=\frac{1}{2}\cdot mv_f^2-\frac{1}{2}\cdot mv_i^2](https://tex.z-dn.net/?f=F_x%5Ccdot%20x%3D%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20mv_f%5E2-%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20mv_i%5E2)
![6.556\times 2.7=\frac{1}{2}\cdot 12\times v_f^2](https://tex.z-dn.net/?f=6.556%5Ctimes%202.7%3D%5Cfrac%7B1%7D%7B2%7D%5Ccdot%2012%5Ctimes%20v_f%5E2)
![v_f^2=\frac{6.556\times 2.7\times 2}{12}](https://tex.z-dn.net/?f=v_f%5E2%3D%5Cfrac%7B6.556%5Ctimes%202.7%5Ctimes%202%7D%7B12%7D)
![v_f^2=2.95](https://tex.z-dn.net/?f=v_f%5E2%3D2.95)
Answer:
mass = 4kg
Explanation:
Kinetic Energy = 1/2 x m x v²
where m = mass and v = velocity
So,
KE = 50
1/2 × m × 5² = 50
1/2 × m × 25 = 50
m = (50 x 2)/25
m = 100/25
m = 4 kg