Changing direction does not cause a change in acceleration.
1 answer:
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Explanation:
If the turntable starts from rest and is set in motion with a constant angular acceleration α. Let is the angular velocity of the turntable. We know that the rate of change of angular velocity is called the angular acceleration of an object. Its formula is given by :
............(1)
Using second equation of kinematics as :
Using equation (1) in above equation
In one revolution, (in 2 revolutions)
Hence, this is the required solution.
Answer:
d) v = 100.2 mph, e) t = 1.25 s, f) -0.0340
Explanation:
d) This is an exercise that we can solve using projectile launch equations, let's start by calculating the time the ball will take to take home-plate
.x = 147 t
t = x / 147
t = 60.5 / 147
t = 0.41156 s
Let's use Pythagoras' theorem to find the speed
v = √ vₓ² + ²
vₓ = dx / dt
vₓ = 147
v_{y} = dy / dt
v_{y} = 4 -16 t
We look for speed for the time of arriving at home
= 4 - 16 0.41156
v_{y} = -2,585 ft / s
v_{y} = -2.585 ft/ s ( 1 mile /5280 foot) (3600s/1h)
Let's calculate the speed
v = √ (147² + 2,585²)
v = 147.02 ft / s
v = 147.0 ft/s (1 mile/5280 feet)(3600s/1h)
v = 100.2 mph
e) the time it takes for the ball to reach the floor and = 0 foot
y = 5 + 4 t - 16 t²
0 = 5 + 4t - 16t²
t² –t / 4 -5/4 = 0
t² -0.25 t -1.25 = 0
We solve the equation and second degree
t = [0.25 ±√(0.25² + 4 1.25)] / 2
t = [0.25 ± 2.25] / 2
t₁ = 1.25 s
t₂ = -1 s
The positive time is correct
t = 1.25 s
f) The angle of speed when the ball passes home
tan θ = / vₓ
θ = tan⁺¹ (v_{y} / vx)
The distance x is given in the exercise
x = 60.5 foot
vₓ = 147 foot / s
The speed y is t = 1.25 s
v_{y} = 5 + 4 1.25 - 16 1.25²
v_{y} = -15 foot / s
θ = tan⁻¹ (-15/147)
θ = -1,947º = -0,0340 rad