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3 years ago

Define nuclear forces

2 answers:

8 0

Nuclear forces means are the forces that act between two or more nucleons. They bind protons and neutrons into atomic nuclei. The nuclear force is about 10 millions times stronger than the chemical binding that holds atoms together in molecules.

Alex17521 [72]3 years ago

6 0

The nuclear force is a force that acts between the protons and neutrons of atoms.

The nuclear force is a force that binds the protons and neutrons in a nucleus together.This force can exist between protons and protons,neutrons and protons and neutrons and neutrons.This force is what hold the nucleus together.

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Which clue indicates that a speciation has occurred between two populations?

Diano4ka-milaya [45]

I believe the answer is C

4 0

3 years ago

Is 250 000 miles from the earth to the moon" is a qualitative<br> Observation<br> TRUE<br> Or false

lys-0071 [83]

Answer:

True

Explanation:

3 0

3 years ago

Suppose a gliding 2-kg cart bumps into, and sticks to, a stationary 5-kg cart. If the speed of the gliding cart before the colli

Thepotemich [5.8K]

Answer:

Therefore,

Final velocity of the coupled carts after the collision is

$v_{f}=4\ m/s$

Explanation:

Given:

Mass of Glidding Cart = m₁ = 2 kg

Mass of Stationary Cart = m₂ = 5 kg

Initial velocity of Glidding Cart = u₁ = 14 m/s

Initial velocity of Stationary Cart = u₂ = 0 m/s

To Find:

Final velocity of the coupled carts after the collision = $v_{f}=?$

Solution:

Law of Conservation of Momentum:

For a collision occurring between two objects in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

It is denoted by "p" and given by

Momentum = p = mass × velocity

Hence by law of Conservation of Momentum we hame

Momentum before collision = Momentum after collision

Here after collision both are stuck together so both will have same final velocity,

$m_{1}\times u_{1}+m_{2}\times u_{2}=(m_{1}+m_{2})\times v_{f}$

Substituting the values we get

$2\times 14 + 5\times 0 =(2+5)\times v_{f}$

$v_{f}=\dfrac{28}{7}=4\ m/s$

Therefore,

Final velocity of the coupled carts after the collision is

$v_{f}=4\ m/s$

8 0

3 years ago

A magnet is located above circular current. What is the direction of the magnetic force on the magnet

Jobisdone [24]

The magnet is attracted to the ring since the north pole of the current loop is above the ring and the south pole is below the ring.

3 0

3 years ago

An antiproton is identical to a proton except it has the opposite charge, −e. To study antiprotons, they must be confined in an

Alex_Xolod [135]

Answer:

d = 1.13*10^{-4}m = 0.113mm

Explanation:

To find the minimum diameter, that allow to antiproton circulate in the chamber without touching the walls, you use the following formula for the radius of the trajectory of a charged particle in a constant magnetic field.

$r=\frac{mv}{qB}$ (1)

r: radius of the trajectory

m: mass of the antiproton = 9.1*10-31 kg

v: velocity of the antiproton = 4.0*10^4 m/s

B: magnitude of the magnetic field = 4.0mT = 4.0*10^-3 T

q: charge of the antiproton = +1.6*10^{-19}C

You replace the values of the parameters in (1):

$r=\frac{(9.1*10^{-31}kg)(4.0*10^4m/s)}{(1.6*10^{-19}C)(4.0*10^{-3}T)}\\\\r=5.68*10^{-5}m$

Then, the diameter of the chamber must be, at least:

d=2r = 2(5.68*10^-5) = 1.13*10^{-4}m = 0.113mm

6 0

4 years ago