Answer:
0.34 sec
Explanation:
Low point of spring ( length of stretched spring ) = 5.8 cm
midpoint of spring = 5.8 / 2 = 2.9 cm
Determine the oscillation period
at equilibrum condition
Kx = Mg
g= 9.8 m/s^2
x = 2.9 * 10^-2 m
k / m = 9.8 / ( 2.9 * 10^-2 ) = 337.93
note : w =
=
= 18.38 rad/sec
Period of oscillation = 
= 0.34 sec
Answer:
First one, third one, and fourth one
Answer:
The magnetic field at a distance of 19.8 cm from the wire is 1.591 mT
Explanation:
Given;
first magnetic field at first distance, B₁ = 2.50 mT
first distance, r₁ = 12.6 cm = 0.126 m
Second magnetic field at Second distance, B₂ = ?
Second distance, r₂ = ?
Magnetic field for a straight wire is given as;

Where:
μ is permeability
B is magnetic field
I is current flowing in the wire
r distance to the wire

Therefore, the magnetic field at a distance of 19.8 cm from the wire is 1.591 mT
Answer:
atoms form bonds by donating, accepting or sharing electrons with other atoms in order to complete their valence shell electrons
hence , C. Bonding gives an atom the same number of protons as a noble gas.
Explanation:
i hope it helped
<span>160 Joules
For this problem, we can ignore the vertical component of the applied force and focus on only the horizontal component of 80 N and since work is defined as force over distance, let's multiply the force by the distance:
80 N * 2.0 m = 160 Nm = 160 kg*m^2/s^2 = 160 Joules.
So the cart has a final kinetic energy of 160 Joules.</span>