Answer:
(a) 1.11sec
(b) 14.37m/s
(c) 31.78m
Explanation:
U = 18m/s, A = 37°, g = 9.8m/s^2
(a) t = UsinA/g = 18sin37°/9.8 = 18×0.6018/9.8 = 1.11sec
(b) Ux = UcosA = 18cos37° = 18×0.7986 = 14.37m/s
(c) R = U^2sin2A/g = 18^2sin2(37°)/9.8 = 324sin74°/9.8 = 324×0.9613/9.8 = 31.78m
Answer:
Yes, this is according to the Newton's first law of motion.
Neither its direction nor its velocity changes during this course of motion.
Explanation:
Yes, it is very well in accordance with Newton's first law of motion for a body with no force acting on it and it travels with a non-zero velocity.
During such a condition the object will have a constant velocity in a certain direction throughout its motion. Neither its direction nor its velocity changes during this course of motion.
Answer:
this answer your question
Answer:
The correct answer is V√5
Explanation:
Let V be the velocity of the satellite orbiting at radius r.
Let V(5r) be the velocity of the satellite orbiting at radius 5r.
Recall:
Escape velocity is given by:
V = √(2gr)
Where V is the escape velocity
g is the acceleration due to gravity
r is the radius of the earth.
With the above equation, we can obtain the answer to the question as follow:
V = √(2gr)
V(5r) = √(2g5r)
Next, we'll obtained the ratio of V(5r) to V as shown below
V(5r) : V => V(5r)/V
V(5r)/V = √(2g5r) / √(2gr)
V(5r)/V = √5
Cross multiply
V(5r) = V√5
From the above illustration, we can see that when the satellite is moved to 3r, then the expression for the velocity will be V√5
