Answer:
v = 2.928 10³ m / s
Explanation:
For this exercise we use Newton's second law where the force is the gravitational pull force
F = ma
a = F / m
Acceleration is
a = dv / dt
a = dv / dr dr / dt
a = dv / dr v
v dv = a dr
We substitute
v dv = a dr
∫ v dv = 1 / m G m M ∫ 1 / r² dr
We integrate
½ v² = G M (-1 / r)
We evaluate from the lower limit v = 0 for r = R m to the upper limit v = v for r = R + 2.73 10³, where R is the radius of Saturn's moon
v² = 2G M (- 1 / R +2.73 10³+ 1 / R)
We calculate
v² = 2 6,674 10⁻¹¹ 1.10 10²¹ (10⁻³ / 5.61 - 10⁻³ /(5.61 + 2.73))
v² = 14.6828 10⁷ (0.1783 -0.1199)
v = √8.5748 10⁶
v = 2.928 10³ m / s
Answer:
Explanation:
From the question we are told that:
Acceleration 
Displacement 
Initial time 
Final Time 
Generally the equation for Velocity of 1.05 travel is mathematically given by
Using Newton's Law of Motion
Generally the equation for Distance traveled before stop is mathematically given by
Generally the equation for Distance to stop is mathematically given by
Since For this Final section
Final velocity 
Initial velocity 
Therefore
Using Newton's Law of Motion
Giving
Therefore
Generally the Total Distance Traveled is mathematically given by
Answer:
Explanation:
Displacement vector along x axes = 4.5 - 2.5 = 2 m
Displacement vector along y axes = 3 - 2 = 1 m
Displacement vector along z axis = 3.5- 4 = - 0.5 m
Displacement vector = 2 i + j - 0.5 k m
Try looking at it like this and I'll bet you'll get it:
You're running through the boxcar on the train at 112 meters per second, but you notice that you're moving along the tracks at 210 meters per second. What must be the speed of the train along the tracks ?
