Effect of looping the loop
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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
Answer:
The electric field is
Explanation:
From the question we are told that
The radius of the inner sphere is
The radius of the outer sphere is
The charge on the inner sphere is
The charge on the outer sphere is
The position from the origin is
Generally the electric field is mathematically represented as
The reason for using for the calculation is due to the fact that the position considered is greater than the
but less than
Here k is the Coulomb constant with value
So