Using the theorem of kinetic energy
1/2mVf² - 1/2mVi²= WF + Wp, Wp=0
WF = F. AB, AB=5m and F= 40N, m=20kg
so the final kinetic is KEf= 1/2mVf² = WF =<span>F. AB= 40*5=200J
</span>
the final velocity is 1/2mVf² <span>=200, implies Vf= sqrt(20)=2sqrt(5)m/s</span>
Answer:
0.86 m
Explanation:
q₁ = magnitude of positive charge = 5 x 10⁻⁶ C
q₂ = magnitude of negative charge = 3 x 10⁻⁶ C
r = distance between the two charges = 0.250 m
d = distance of the location of third charge from negative charge
q = magnitude of charge on third charge
Using equilibrium of electric force on third charge



d = 0.86 m
<span>D. unbalanced force, motion, and movement in the same direction of the applied force
Hope this helps!</span>
Answer:
196J
Explanation:
Given parameters:
Mass of skateboard = 1kg
Height off the ground = 20m
Unknown:
Potential energy = ?
Solution:
The potential energy is the energy due to the position of a body above the ground.
It is mathematically expressed as:
Potential energy = mass x acceleration due gravity x height
Potential energy = 1 x 9.8 x 20 = 196J
Answer:
y = x tan θ - (g / 2v₀² cos² θ) x²
Explanation:
An equation is called a general formula that relates the position on the x-axis and the height on the body's axis.
Let's write the position on each axis
X axis. No acceleration
x = v₀ₓ t
Y Axis. There is the acceleration of gravity
y =
t - ½ g t²
Let's clear the time in the first equation and substitute in the second
y = v₀ sin θ (x / v₀ₓ) - ½ g (x / v₀ₓ)²
y = v₀ sin θ / v₀ cos θ x - ½ g x² / v₀² cos² θ
y = x tan θ - (g / 2v₀² cos² θ) x²
This is the trajectory equation in projectile launching