<span>Given:
3,500 kilometers
Find:</span>
Years for two continents to collide = ?
<span>Solution:
We know that </span>typical motions of one plate relative to another
are 1 centimeter per year.
So first, we convert 3,500 km to cm.<span>
</span><span>
</span>
The solution would be like this for this specific problem:
1 km = 100,000 cm
3,500 km x 100,000 = 350,000,000 cm
Since we know that 1 cm = 1 year, then that means
350,000,000 cm is equivalent to 350,000,000 years.
Therefore, it would take 350 million years for two continents
that are 3500 kilometers apart to collide.
<span>
To add, </span>a phenomenon of the plate tectonics of Earth that occurs at
convergent boundaries is called the continental collision.
Answer:
a) 
b) 
Explanation:
Searching the missed information we have:
E: is the energy emitted in the plutonium decay = 8.40x10⁻¹³ J
m(⁴He): is the mass of the helium nucleus = 6.68x10⁻²⁷ kg
m(²³⁵U): is the mass of the helium U-235 nucleus = 3.92x10⁻²⁵ kg
a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:
Linear momentum:
Since the plutonium nucleus is originally at rest, 
:
(1)
Kinetic Energy:
(2)
By entering equation (1) into (2) we have:
Solving the above equation for 
we have:
And by entering that value into equation (1):
The minus sign means that the helium-4 nucleus is moving in the opposite direction to the uranium-235 nucleus.
b) Now, the kinetic energy of each nucleus is:
For He-4:
For U-235:
I hope it helps you!
230 Newton
Electric charge consists of two types i.e. positively electric charge and negatively electric charge.There was a famous scientist who investigated about this charges. His name is Coulomb and succeeded in formulating the force of attraction or repulsion between two charges i.e. :
F = electric force (N)
k = electric constant (N m² / C²)
q = electric charge (C)
r = distance between charges (m)
The value of k in a vacuum = 9 x 10⁹ (N m² / C²)
F = k(q1 q2)/ r^2
Distance between protons = d = 10⁻¹⁵ m
charge of proton = q = 1.6 × 10⁻¹⁹ C
Here q1=q2
electric force = F =230N
Coulomb's Law. Two protons in an atomic nucleus are typically separated by a distance of 2×10−15m. The electric repulsive force between the protons is huge, but the attractive nuclear force is even stronger and keeps the nucleus from bursting apart.
2 Nuclei and the Need for an Attractive Nuclear Force. The Coulomb force also acts within atomic nucleii, whose characteristic dimension is 10 m, which is called a fermi. There are two protons in a He nucleus, which repel each other because of the Coulomb force.
Find more about electric force of repulsion between nuclear protons
brainly.com/question/8404637
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