Answer:
<em>The velocity of the carts after the event is 1 m/s</em>
Explanation:
<u>Law Of Conservation Of Linear Momentum
</u>
The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is
P=mv.
If we have a system of bodies, then the total momentum is the sum of the individual momentums:

If a collision occurs and the velocities change to v', the final momentum is:

Since the total momentum is conserved, then:
P = P'
In a system of two masses, the equation simplifies to:

If both masses stick together after the collision at a common speed v', then:

The common velocity after this situation is:

The m1=2 kg cart is moving to the right at v1=5 m/s. It collides with an m2= 8 kg cart at rest (v2=0). Knowing they stick together after the collision, the common speed is:

The velocity of the carts after the event is 1 m/s
Answer:
<em>It matters because crystalline and amorphous materials have different properties. The arrange affects the melting point (defined in crystals and a larger range in amorphous) and shape (geometrical in crystals, no geometrical in amorphous). </em>
Explanation:
The particles that compose a solid material are held in place by strong tractive forces between them when we analyze solids we consider the position of the atoms (molecules or ions) rather than their motion (which is important in liquids and gases). This positioning can be arranged in two general ways:
- Crystalline solids have internal structures that in turn lead to distinctive flat surfaces or face, these faces intersect at angles that are characteristic of the substance, crystals tend to have sharp, well defined and high melting points because of the same distance from the same number and type of neighbors. They generally have geometric shapes, some examples are diamonds, metals, salts.
- Amorphous solids produce irregular or curved surfaces when broken and they have poorly defined patterns when exposed to x rays because of their irregular array. In contrast with crystal solids, amorphous solids soften over a wide temperature range due to the different amounts of thermal energy needed to overcome different interactions. Some examples of these solids are gels, plastics, and some polymers.
I hope you find this information useful and interesting! Good luck!
The correct answer is remain the same because if the particles and temperature remain constant, when the pressure increases the temperature is not increasing so it would just stay the same
Answer:
Explanation:
given,
A woman can see object between 45 cm and 155 cm
glasses is 2 cm from the eye
to able to read a book distance = 25 cm
For the object distant apart



hence, D = - 0.653 m
For the object at the near point



hence, D = 2.02 m