Stars are formed in <u>nebulas</u>, interstellar clouds of dust and gas.
Draw a diagram to illustrate the problem as shown below.
The vertical component of the launch velocity is
v = (8.5 m/s)*sin30° = 4.25 m/s
The horizontal component of the launch velocity is
8.5*cos30° = 7.361 m/s
Assume that aerodynamic resistance may be ignored.
Because the horizontal distance traveled is 19 m, the time of travel is
t = 19/7.361 = 2.581 s
The downward vertical travel is modeled by
h = (-4.25 m/s)*(2.581 s) + 0.5*(9.8 m/s²)*(2.581 s)²
= 21.675 m
Answer: The height is 21.7 m (nearest tenth)
Answer:
The x-component of
is 56.148 newtons.
Explanation:
From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:
(1)
Where:
,
,
- External forces exerted on the ring, measured in newtons.
- Vector zero, measured in newtons.
If we know that
,
,
and
, then we construct the following system of linear equations:
(2)
(3)
The solution of this system is:
, 
The x-component of
is 56.148 newtons.
Complete question is:
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.
Answer:
(V_A) = 31.32 m/s
Explanation:
We are given;
car's mass, m = 1200 kg
h_A = 100 m
h_B = 150 m
v_B = 0 m/s
From law of conservation of energy,
the distance from point A to B is;
h = 150m - 100 m = 50 m
From Newton's equations of motion;
v² = u² + 2gh
Thus;
(V_B)² = (V_A)² + (-2gh)
(negative next to g because it's going against gravity)
Thus;
(V_B)² = (V_A)² - (2gh)
Plugging in the relevant values;
0² = (V_A)² - 2(9.81 × 50)
(V_A) = √981
(V_A) = 31.32 m/s