Answer:
The compression of the spring is 24.6 cm
Explanation:
magnitude of the charge on the left, q₁ = 4.6 x 10⁻⁷ C
magnitude of the charge on the right, q₂ = 7.5 x 10⁻⁷ C
distance between the two charges, r = 3 cm = 0.03 m
spring constant, k = 14 N/m
The attractive force between the two charges is calculated using Coulomb's law;
The extension of the spring is calculated as follows;
F = kx
x = F/k
x = 3.45 / 14
x = 0.246 m
x = 24.6 cm
The compression of the spring is 24.6 cm
Answer:
Velocity of Object with 2 kg= 3.390 m/s
Velocity of Object with 3 kg= 3.404 m/s
Explanation:
From the picture, it can be seen that object B is initially at rest while object A is travelling at a speed of 5m/s. After the collision, Object A moves at an angle of 65 degrees while object B moves at an angle of 37 degrees.
We also know that momentum of a closed system is conserved.
Initial momentum along the x-axis = 2*5.5 = 11
Initial momentum along y-axis = 0
Final momentum along x-axis= a*Cos(65)*2 +b*Cos(37) *3= 11 (a is the velocity of object A of 2 kg after collision where as b is the velocity of object B of 3 kg after collision. velocity is multiplied by cosines of the angle from x axis to give the horizontal component of the velocities).
Final momentum along y-axis = a*Sin(65)*2 - b*Sin(37)*3 =0 (We can see that vertical components of velocity are opposite in direction to each other)
Solve both the equations simultaneously for a and b.
Explanation:
You can solve for volume using radius or diameter.
Sphere Volume = 4/3 • π • r³ = ( π •d³)/6
We're given the diameter so let's use that.
Volume = PI * d^3 / 6
Volume = 3.14159 * 3.0^3 / 6
Volume = 3.14159 * 9 / 2
Volume = 14.137 cubic centimeters
