I'm not positive but if I'm reading the question right it would be the big bang sorry if I'm wrong
Energy in a spring:
E = 0.5 * k * x²
k spring constant = 800 n/m
x stretch of the spring = 5 cm = 0.05 m
E = 0.5 * 800 * 0.05² = 1
It can be noted that the directions west and north are perpendicular to each other. hence we can use the right angle formula for calculating total distance in these directions. Also South is just negative north. This technique is called euclidean distance norm. It is the basis of Cartesian coordinate system.
Total distance = 3.22 N + 4.75 W +1.90 S
= 1.32 N + 4.75 W
D = 
D = 4.93 Km
direction = atan(N/W) = 15.53 deg north of west
Answer: m∠P ≈ 46,42°
because using the law of sines in ΔPQR
=> sin 75°/ 4 = sin P/3
so ur friend is wrong due to confusion between edges
+) we have: sin 75°/4 = sin P/3
=> sin P = sin 75°/4 . 3 = (3√6 + 3√2)/16
=> m∠P ≈ 46,42°
Explanation:
Answer:
a) E = -4 10² N / C
, b) x = 0.093 m, c) a = 10.31 m / s², θ=-71.9⁰
Explanation:
For that exercise we use Newton's second Law, in the attached we can see a free body diagram of the ball
X axis
-
= m a
Axis y
- W = 0
Initially the system is in equilibrium, so zero acceleration
Fe =
T_{y} = W
Let us search with trigonometry the components of the tendency
cos θ = T_{y} / T
sin θ =
/ T
T_{y} = cos θ
= T sin θ
We replace
q E = T sin θ
mg = T cosθ
a) the electric force is
= q E
E =
/ q
E = -0.032 / 80 10⁻⁶
E = -4 10² N / C
b) the distance to this point can be found by dividing the two equations
q E / mg = tan θ
θ = tan⁻¹ qE / mg
Let's calculate
θ = tan⁻¹ (80 10⁻⁶ 4 10² / 0.01 9.8)
θ = tan⁻¹ 0.3265
θ = 18
⁰
sin 18 = x/0.30
x =0.30 sin 18
x = 0.093 m
c) The rope is cut, two forces remain acting on the ball, on the x-axis the electric force and on the axis and the force gravitations
X axis
= m aₓ
aₓ = q E / m
aₓ = 80 10⁻⁶ 4 10² / 0.01
aₓ = 3.2 m / s²
Axis y
W = m
a_{y} = g
a_{y} = 9.8 m/s²
The total acceleration is can be found using Pythagoras' theorem
a = √ aₓ² + a_{y}²
a = √ 3.2² + 9.8²
a = 10.31 m / s²
The Angle meet him with trigonometry
tan θ = a_{y} / aₓ
θ = tan⁻¹ a_{y} / aₓ
θ = tan⁻¹ (-9.8) / 3.2
θ = -71.9⁰
Movement is two-dimensional type with acceleration in both axes