Force equals mass*distance
F = ma
Given m = 10 kg, F = 30 N
30 = 10a
30/10 = a
3 = a
The wagon's acceleration is 3 m/s^2
Answer:
1. <u>F = ma</u> <em>F = 0.2kg * 20m/s² = 4Kg * m/s² =</em> 4N
2. <u>F = ma</u> <em>F - 18Kg * 3m/s² = 54Kg * m/s² =</em> 54N
3. <u>F = ma</u> <em>F = 0.025Kg * 5m/s² =</em> 0.125N
4. <u>F = ma</u> <em>F = 50Kg * 4m/s² =</em> 200N
5. <u>F = ma</u> <em>F = 70Kg * 4m/s² =</em> 280N
6. <u>F = ma</u> <em>F = 9Kg * 9.8m/s² =</em> 88.2N
Explanation:
Hope this helps ! ^^
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
Imagine an object is moving in one dimension on a number line, and for this we'll say that the numbers on the line are a metre apart. If the object moves from 2 m to 7 m, the change in position is 7-2=+5 metres. But if the object moves back from 7 m to 2 m, the change in position is 2-7=-5 metres. since

, and time is always positive, velocity will be positive in one direction and negative in the other direction.