A force of 25 newtons moves a box a distance of 4 meters in 5 seconds.
2 answers:
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Answer:
The gravitational force between m₁ and m₂, is approximately 1.06789 × 10⁻⁶ N
Explanation:
The details of the given masses having gravitational attractive force between them are;
m₁ = 20 kg, r₁ = 10 cm = 0.1 m, m₂ = 50 kg, and r₂ = 15 cm = 0.15 m
The gravitational force between m₁ and m₂ is given by Newton's Law of gravitation as follows;
Where;
F = The gravitational force between m₁ and m₂
G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²
r₂ = 0.1 m + 0.15 m = 0.25 m
Therefore, we have;
The gravitational force between m₁ and m₂, F ≈ 1.06789 × 10⁻⁶ N
1) Distance down the hill: 1752 ft (534 m)
2) Time of flight of the shell: 12.9 s
3) Final speed: 326.8 ft/s (99.6 m/s)
Explanation:
1)
The motion of the shell is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.
The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:
(1)
where:
is the initial vertical velocity of the shell, with
and
is the acceleration of gravity
At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation
where is the initial horizontal velocity of the shell.
We can re-write this last equation as
(1b)
And substituting into (1),
(2)
where we have choosen the top of the hill (starting position of the shell) as origin (0,0).
We also know that the hill goes down with a slope of from the horizontal, so we can write the position (x,y) of the hill as
(3)
Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):
Substituting (1b) into this equation,
Which has 2 solutions:
x = 0 (origin)
and
So, the distance d down the hill at which the shell strikes the hill is
2)
In order to find how long the mortar shell remain in the air, we can use the equation:
where:
x = 1322 ft is the final position of the shell when it strikes the hill
is the initial velocity of the shell
is the angle of projection of the shell
Substituting these values into the equation, we find the time of flight of the shell:
3)
In order to find the final speed of the shell, we have to compute its horizontal and vertical velocity first.
The horizontal component of the velocity is constant and it is
Instead, the vertical component of the velocity is given by
And substituting at t = 12.9 s (time at which the shell strikes the hill),
Therefore, the final speed of the shell is:
Learn more about projectile motion:
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