The distance it falls is given by
x = (1/2)at^2
where a = acceleration due to gravity = 9.8 m/s^2
x = (1/2)(9.8)(18)^2
x = 1587.6 m
The answer is 1587.6 meters
Answer:
- When an object experiences acceleration to the left, the net force acting on this object will also be to the left.
- If the mass of the object was doubled, it would experience an acceleration of half the magnitude
Explanation:
When an object experiences acceleration to the left, the net force acting on this object will also be to the left.
From Newton's second law of motion, the acceleration of the object is given as;
a = ∑F / m
a = -F / m
The negative value of "a" indicates acceleration to the left
where;
∑F is the net force on the object
m is the mass of the object
At a constant force, F = ma ⇒ m₁a₁ = m₂a₂
If the mass of the object was doubled, m₂ = 2m₁
a₂ = (m₁a₁) / (m₂)
a₂ = (m₁a₁) / (2m₁)
a₂ = ¹/₂(a₁)
Therefore, the following can be deduced from the acceleration of this object;
- When an object experiences acceleration to the left, the net force acting on this object will also be to the left.
- If the mass of the object was doubled, it would experience an acceleration of half the magnitude
Answer:
C. Why you must push harder to move a car farther.
Explanation:
Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.
Mathematically, it is given by the formula;
Hence, Newton's 2nd Law explains why you must push harder to move a car farther because of its mass. Thus, it is important to increase the force that the engine provides and decrease the mass of the car.
Answer:
Learning the formula.multiply mass accelebrations.the force(F)required to move an object of mass(M) with an acceleration (a) is given by the formula F = m x a.so, force = mass multiplied by accelebration.
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