Answer:
Final velocity = 7.677 m/s
KE before crash = 202300 J
KE after crash = 182,702.62 J
Explanation:
We are given;
m1 = 1400 kg
m2 = 4700 kg
u1 = 17 m/s
u2 = 0 m/s
Using formula for inelastic collision, we have;
m1•u1 + m2•u2 = (m1 + m2)v
Where v is final velocity after collision.
Plugging in the relevant values;
(1400 × 17) + (4700 × 0) = (1400 + 1700)v
23800 = 3100v
v = 23800/3100
v = 7.677 m/s
Kinetic energy before crash = ½ × 1400 × 17² = 202300 J
Kinetic energy after crash = ½(1400 + 1700) × 7.677² = 182,702.62 J
The term “electric field” refers to the physical field that surrounds electrically charged particles and acts to either attract or repel all other charged particles in the field (also known as an E-field).
It can also refer to the physical field surrounding a system of charged particles. Electric fields are composed of electric charges and time-varying electric currents.
Both electric and magnetic fields are manifestations of the electromagnetic field, one of the four fundamental interactions (sometimes known as forces) of nature.
Electrical technology makes use of electric fields, which are significant in many branches of physics.
For instance, in atomic physics and chemistry, the electric field acts as an attractive force to hold atoms’ atomic nuclei and electrons together. It is also the force that causes atoms to chemically link together to form molecules.
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Answer:
that is cool and i have one interesting fact
Explanation: North Korea and Cuba are the only places you can't buy Coca-Cola
The electric potential energy of the pair of charges when the second charge is at point b is 7.3 x 10⁻⁸ J.
<h3>
Electric potential energy</h3>
When work is done on a positive test charge to move it from one location to another, potential energy increases and electric potential increases.
The electric potential energy between the charges when the second charge is at point b is calculated as follows;
ΔU = -w
Ui - Uf = w
Uf = Ui - w
where;
Uf is the final potential energy
Ui is the initial potential energy
w is the work done by the force
Uf = 5.4 x 10⁻⁸ J - (-1.9 x 10⁻⁸J)
Uf = 5.4 x 10⁻⁸ J + 1.9 x 10⁻⁸ J
Uf = 7.3 x 10⁻⁸ J
Thus, the electric potential energy of the pair of charges when the second charge is at point b is 7.3 x 10⁻⁸ J.
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