Data:
m (mass) = 1 Kg
s (speed) = 3 m/s
Kinetic energy = ? (Joule)
Formula (Kinetic energy)

Solving:




Answer:
60 m/s
Explanation:
From the law of conservation of energy,
The kinetic energy of the plane = Energy of store in the spring when the plane lands.
1/2mv² = 1/2ke²
making v the subject of the equation.
v = √(ke²/m).................... Equation 1
Where v = the plane landing speed, k = spring constant, e = extension. m = mass of the plane.
Given: m = 15000 kg, k = 60000 N/m, e = 30 m.
Substitute into equation 1
v = √(60000×30²/15000)
v = √(4×900)
v = √(3600)
v = 60 m/s.
Hence the plane's landing speed = 60 m/s
Answer:
The center of mass of the two-ball system is 7.05 m above ground.
Explanation:
<u>Motion of 0.50 kg ball:</u>
Initial speed, u = 0 m/s
Time = 2 s
Acceleration = 9.81 m/s²
Initial height = 25 m
Substituting in equation s = ut + 0.5 at²
s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m
Height above ground = 25 - 19.62 = 5.38 m
<u>Motion of 0.25 kg ball:</u>
Initial speed, u = 15 m/s
Time = 2 s
Acceleration = -9.81 m/s²
Substituting in equation s = ut + 0.5 at²
s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m
Height above ground = 10.38 m
We have equation for center of gravity

m₁ = 0.50 kg
x₁ = 5.38 m
m₂ = 0.25 kg
x₂ = 10.38 m
Substituting

The center of mass of the two-ball system is 7.05 m above ground.
Nacl is the answer hahahahaah ha
Answer:
The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.
Explanation:
We can find the drift speed by using the following equation:
Where:
I: is the current = 4.50 A
n: is the number of electrons
q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C
A: is the cross-sectional area = 2.20x10⁻⁶ m²
We need to find the number of electrons:
Now, we can find the drift speed:
Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.
I hope it helps you!