Answer:
0.853 m/s
Explanation:
Total energy stored in the spring = Total kinetic energy of the masses.
1/2ke² = 1/2m'v².................... Equation 1
Where k = spring constant of the spring, e = extension, m' = total mass, v = speed of the masses.
make v the subject of the equation,
v = e[√(k/m')].................... Equation 2
Given: e = 39 cm = 0.39 m, m' = 0.4+0.4 = 0.8 kg, k = 1.75 N/cm = 175 N/m.
Substitute into equation 2
v = 0.39[√(1.75/0.8)
v = 0.39[2.1875]
v = 0.853 m/s
Hence the speed of each mass = 0.853 m/s
Explanation:
Magnet: It has two poles: South pole and North pole.
Magnetic field lines are stronger near the poles of the magnet.
Same poles repel each other. There is a magnetic force of repulsion between the same poles. North- North poles repel each other.
Unlike poles attract each other. There is magnetic force of attraction between the opposite poles. South- North poles attract each other.
Mono poles cannot exist.
From the given statements, the magnetic poles are described by:
A north pole must exist with a south pole.
Two south poles placed near each other will repel each other.
A north pole and a south pole placed near each other will attract each other.
I will answer both versions assuming what you want to know is the distance it travels up from and over the ground. and how long until it reaches space. 540 meters per second up and over. to reach space which is 100km above sea level, it would take about 5400 minutes
Answer:
3.8 x 10^10 m/s^2
Explanation:
Charge, q = 2 e = 2 x 1.6 x 10^-19 C = 3.2 x 10^-19 C
Electric field strength, E = 790 N/C
mass of helium nucleus, m = 6.645 x 10^-27 Kg
the force due to electric filed on a charge particle is given by
F = q x E
Where, q be the charge on the charged particle and E be the strength of electric field.
By substituting the values
F = 3.2 x 10^-19 x 790
F = 2528 x 10^-19 N
According to the Newton's second law
F = m x a
Where, me be the mass and a be the acceleration
By substituting the values
a = 3.8 x 10^10 m/s^2
Explanation:
I am not sure about this question sry but u can try asking a tutor u don't need to use any points
