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borishaifa [10]
3 years ago
14

Why do we add the masses together after that inelastic collision?

Physics
1 answer:
fomenos3 years ago
5 0

Answer:An inelastic collision is one in which the internal kinetic energy changes (it is not conserved). A collision in which the objects stick together is sometimes called perfectly inelastic because it reduces internal kinetic energy more than does any other type of inelastic collision.People sometimes think that objects must stick together in an inelastic collision. However, objects only stick together during a perfectly inelastic collision. Objects may also bounce off each other or explode apart, and the collision is still considered inelastic as long as kinetic energy is not conserved.

hope this helps have a nice day❤️

Explanation:

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A train accelerates from 30 km/h to 45 km/h in 15.0 second. Find its acceleration and the distance it travels during this time
Bad White [126]

Answer:

a. Acceleration, a = 0.28 m/s²

b. Distance, S = 156 meters

Explanation:

Given the following data;

Initial velocity = 30 km/h

Final velocity = 45 km/h

Time = 15 seconds

a. To find the acceleration;

Conversion:

30 km/h to m/s = 30*1000/3600 = 8.33 m/s

45 km/h to m/s = 45*1000/3600 = 12.5 m/s

Mathematically, acceleration is given by the equation;

Acceleration (a) = \frac{final \; velocity  -  initial \; velocity}{time}

Substituting into the equation;

a = \frac{12.5 - 8.3}{15}

a = \frac{4.2}{15}

Acceleration, a = 0.28 m/s²

b. To find the distance travelled, we would use the second equation of motion given by the formula;

S = ut + \frac {1}{2}at^{2}

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

S = 8.3*15 + \frac {1}{2}*(0.28)*15^{2}

S = 124.5 + 0.14*225

S = 124.5 + 31.5

S = 156 meters

8 0
3 years ago
Two 2.0 g plastic buttons each with + 40 nC of charge are placed on a frictionless surface 2.0 cm (measured between centers) on
EleoNora [17]

Answer:

a. There are three potential energy interaction. b. 2.16 m/s c. 2.16 m/s d. 0 m/s

Explanation:

a. There are three potential energy interaction.

Let the charges be q₁ = +40 nC, q₂ = +250 nC and q₃ = + 40 nC and the distances between them be q₁ and q₂ is r, the distance between q₂ and q₃ is r  and the distance between q₁ and q₃ is  r₁ = 2r respectively. So, the potential energies are

U₁ = kq₁q₂/r, U₂ = kq₁q₃/2r and U₃ = kq₂q₃/r

U = U₁ + U₂ + U₃ = kq₁q₂/r +  kq₁q₃/2r + kq₂q₃/r (q₁ = q₃ = q and q₂ = Q)

U = kqQ/r +  kq²/2r + kqQ/r = qk/r(2Q + q/2)

b. To calculate the final speed of the left 2.0 g button, the potential energy = kinetic energy change of the particle.

ΔU = -ΔK

0 - qk(2Q + q/2)/r = -(1/2mv² - 0). Since the final potential at infinity equals zero and the initial kinetic energy is zero.

So qk(2Q + q/2)/r = -1/2mv²

v = √[2qk(2Q + q/2)/mr] where m = 2.0 g r = 2.0 cm

substituting the values for the variables,

v = √[2 × 40 × 10⁻⁹ × 9 × 10⁹(2 × 250 × 10⁻⁹ + 40 × 10⁻⁹/2)/2 × 10⁻³ × 2 × 10⁻²]

v = √[360(500 × 10⁻⁹ + 20 × 10⁻⁹)/2 × 10⁻⁵]

v = √[720(520 × 10⁻⁹)/4 × 10⁻⁵] = 2.16 m/s

c. The final speed of the right 2.0 g button is also 2.16 m/s since we have the same potential energy in the system

d.

Since the net force on the 5.0 g mass is zero due to the mutual repulsion of the charges on the two 2.0 g masses, its acceleration a = 0. Since it starts from rests u = 0, its velocity v = u + at.

Hence,

v = u + at = 0 + 0t = 0 m/s

8 0
4 years ago
When an ideal gas does positive work in its surroundings, which of the ga's quantities increases?
Burka [1]
I think it's B hope it helps
4 0
3 years ago
At take off, a plane flies 100 km north before turning to fly 200 km east. How far is its destination from where the plane took
koban [17]

224 km I did the quiz today

7 0
3 years ago
Read 2 more answers
Which of these statements best proves that electromagnetic forces are weaker than strong nuclear forces? Neutrons do not repel o
musickatia [10]
Answer: <span>Protons do not move out of the nucleus of atoms although they repel each other.

Reasoning:

Remember that protons are particles with positive charge and they held together in the nucleus of the atom which is a tiny tiny region. As you know, like charges repel each other, which means that the protons exert a repulsion force.

That repulsion force tends to destabilize the atom. The only reason why the atom does not explode is because there is a there exists a force of attraction that exceeds the repulsion of the like charged protons. That is the strong nuclear force. Therefore, the strong nuclear forces are stronger than the electromagnetic force that tends to get the protons apart.
</span>
7 0
3 years ago
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