A metal rod A and a metal sphere B, on insulating stands, touch each other. They are originally neutral. A positively charged ro
1 answer:
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Answer:
a) x = v₀ₓ t
, y = t - ½ g t²
Explanation:
This is a projectile launch problem, let's use Newton's second law on each axis
X axis
F = m a
Since there is no acceleration on the x axis, the force on this axis is zero
Y Axis
-W = m
-m g = m
= -g
In this axis the acceleration is the acceleration of gravity
Now we can use science to find the position of the body on each axis
X axis
x = v₀ₓ t + ½ a t²
As the acceleration on this axis is zero
x = v₀ₓ t
Y Axis
y = t + ½
t²
The acceleration on this axis is –g
y = t - ½ g t²
B) to find the maximum value of distance r
r =√ x² + y²
r = √( v₀ₓ² t² + ( t + ½ g t²)²
We can find the maximum value of r using time respect derivatives
dr / dt = 0
0 = ½ 1/√( v₀ₓ² t² + ( t + ½ g t²)² (v₀ₓ² 2t + 2 (
t - ½ g t²)(
- ½ g2t)
We simplify this expression
0 = v₀ₓ² 2t + 2 ( t - ½ g t²) (
- ½ g2t)
-v₀ₓ² t =( t - ½ g t²) (
- ½ g2t)
-v₀ₓ² t = ² t -3/2 gt²
+ 1/2 g² t³
½ g² t²- 3/2 g t =
² + v₀ₓ²
Let's use trigonometry to find go and vox
sin θ = / v₀
cos θ = vox / v₀
= v₀ sin θ
v₀ₓ = v₀ cos θ
We replace
½ g² t² -3/2 g t = v₀ (sin² θ + cos² θ)
g t² - 3v₀ sin θ t = 2 v₀/g
The time is maximum for the angle is zero
Answer:
The pressure is
Explanation:
From the question we are told that
The depth of the swimming pool is
The density of water is
Generally the total pressure exerted on the bottom of the swimming pool is mathematically represented as
Here is the atmospheric pressure with value
So
=>
=>
Apply the Pythagorean theorem to get the resultant velocity:
V =
Given values:
Vx = 57.8m/s
Vy = 5.6m/s
Plug in and solve for V:
V = 58.1m/s
EDIT: Let's get the direction of the resultant velocity as well.
This equation will give the angle of the velocity as measured off of the ground:
θ = tan⁻¹(Vy/Vx)
Again, the given values are:
Vx = 57.8m/s
Vy = 5.6m/s
Plug in the values and solve for the angle θ:
θ = tan⁻¹(5.6/57.8)
θ = 5.5°
The resultant velocity is oriented 5.5° off the ground.