a plane flies 400 km east from city a to city b in 45.0 min and then 980 km south from city b to city c in 1.50 h. (assume the x
1 answer:
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Answer:
No, the farmer is not able to move the mule.
Explanation:
Mass =100 kg
Force=F=800 N
The coefficient between the mule and the ground=
Static friction force,f=
Normal force=N=mg
Static friction force,f=
Using
F<f
Static friction force is greater than applied force.
Therefore , the farmer is not able to move the mule.
Answer:
Height of the rocket be one minute after liftoff is 40.1382 km.
Explanation:
v = velocity of rocket at time t
g = Acceleration due to gravity =
= Constant velocity relative to the rocket = 2,900m/s.
m = Initial mass of the rocket at liftoff = 29000 kg
r = Rate at which fuel is consumed = 170 kg/s
Velocity of the rocket after 1 minute of the liftoff =v
t = 1 minute = 60 seconds'
Substituting all the given values in in the given equation:
Height of the rocket = h
Height of the rocket be one minute after liftoff is 40.1382 km.
Answer:
a) W = 46.8 J and b) v = 3.84 m/s
Explanation:
The energy work theorem states that the work done on the system is equal to the variation of the kinetic energy
W = ΔK = -K₀
a) work is the scalar product of force by distance
W = F . d
Bold indicates vectors. In this case the dog applies a force in the direction of the displacement, so the angle between the force and the displacement is zero, therefore, the scalar product is reduced to the ordinary product.
W = F d cos θ
W = 39.0 1.20 cos 0
W = 46.8 J
b) zero initial kinetic language because the package is stopped
W - =
-K₀
W - fr d= ½ m v² - 0
W - μ N d = ½ m v
on the horizontal surface using Newton's second law
N-W = 0
N = W = mg
W - μ mg d = ½ m v
v² = (W -μ mg d) 2/m
v = √(W -μ mg d) 2/m
v = √[(46.8 - 0.30 4.30 9.8 1.20) 2/4.3 ]
v = √(31.63 2/4.3)
v = 3.84 m/s