How does the force of gravity between two bodies change when the distance between them doubles? 1. unable to determine; the mass
1 answer:
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Answer:
F = 63N
Explanation:
M= 1.5kg , t= 2s, r = (2t + 10)m and
Θ = (1.5t² - 6t).
magnitude of the resultant force acting on 1.5kg = ?
Force acting on the mass =
∑Fr =MAr
Fr = m(∇r² - rθ²) ..........equation (i)
∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)
The horizontal path is defined as
r = (2t + 10)
dr/dt = 2, d²r/dt² = 0
Angle Θ is defined by
θ = (1.5t² - 6t)
dθ/dt = 3t, d²θ/dt² = 3
at t = 2
r = (2t + 10) = (2*(2) +10) = 14
but dr/dt = 2m/s and d²r/dt² = 0m/s
θ = (1.5(2)² - 6(2) ) = -6rads
dθ/dt =3(2) - 6 = 0rads
d²θ/dt = 3rad/s²
substituting equation i into equation ii,
Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)
∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]
∑F = 1.5(14*3+0) = 63N
F = √(Fr² +FΘ²) = √(0² + 63²) = 63N
Answer:
A)
D) (Speed magnitude)
α = 22.49° (Speed direction above the horizontal)
Explanation:
Conceptual analysis:
To solve this problem we consider the following concepts:
1) The projectile in its movement describes a curved line called a parabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane.
The initial velocity (Vo) is tangent to the trajectory at the initial point and can be broken down into two components, one vertical (Voy) and one horizontal(Vox):
Formula (1)
Formula (2)
Where:
Vo: Initial velocity in m/s
: Initial angle above the horizontal in grades
2) The formula to calculate its velocity at any vertical position(y) is as follows:
Formula (3)
Where:
: Final speed component in vertical direction in m/s
: Initial speed component in vertical direction in m/s
g: acceleration due to gravity in m/s2
3) The formulas to calculate the projectile velocity components at any time (t) are:
Because the movement is uniform in the x direction (constant speed)
Formula (4) Because the movement is uniformly accelerated in the y direction
Known information:
We know the following data:
= 34.5º above the horizontal
Development of the problem:
Initial speed components(Vox, Voy), (Formula (1), Formula (2)
A) Maximum height (h):
When the projectile reaches its maximum height (h) ,the speed component Vy = 0, then, we replace this value in the Formula (3):
D) Speed of the projectile 1.50s after firing
We replace t=1.5 s in the formula(4)
(Speed magnitude)
α = 22.49° (Speed direction above the horizontal)