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FromTheMoon [43]

3 years ago

5

How is an emulsion different from other colloids?

1 answer:

Usimov [2.4K]3 years ago

8 0

Answer:

The difference between colloid and emulsion is that a colloid can form when any state of matter (solid, liquid or gas) combine with a liquid whereas an emulsion has two liquid components which are immiscible with each other

Explanation:

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In an isolated system, a red ball of mass m moves to the right with speed v. It strikes a green ball, of

ella [17]

The answer is D. To the left with speed 2v

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3 years ago

Heavy crate sits at rest on the floor of a warehouse. you push on the crate with a force of 400 n, and it doesn't budge. what is

maks197457 [2]

Heavy crate sits at rest on the floor of a warehouse. you push on the crate with a force of 400 N, and it doesn't budge. The magnitude of the friction force on the crate in Newton is 400N

This is due to Friction force, which is defined as the resisting force that acts on a body when it is at rest (Static friction) or when it is in motion (Kinetic friction).

When a force is applied on a stationary body, the force of static friction starts to act on the body which prevents any relative motion between the object and surface. The magnitude of friction increases up to μsN, where μs is the coefficient of static friction. As the crate didn't budge, it means the amount of force applied was less than μsN. Hence the force applied was canceled by an equal and opposite amount of frictional force which was equal to 400N.

Learn more about frictional force here

brainly.com/question/1714663

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8 0

2 years ago

Can somone pls help me??!! i’m very stuck

natima [27]

Answer:

graph b

Explanation:

graph a isnt constant

graph c is slower (less steep slope)

graph d is slowing down (starts high ends low)

5 0

3 years ago

A car traveling 25 m/s to the north collides with a car traveling 15 m/s to the south. Each car has a mass of 1100 kg. If the sy

kirill [66]

conservation of momentum ... that before = that after ... change zero

8 0

3 years ago

Read 2 more answers

A 100 g ball collides elastically with a 300 g ball that is at rest. If the 100 g ball was traveling

sammy [17]

Answer:

The magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

Explanation:

We can find the velocity of the two balls after the collision by conservation of linear momentum and energy:

$P_{1} = P_{2}$

$m_{1}v_{1_{i}} + m_{2}v_{2_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

Where:

m₁: is the mass of the ball 1 = 100 g = 0.1 kg

m₂: is the mass of the ball 2 = 300 g = 0.3 kg

$v_{1_{i}}$ : is the initial velocity of the ball 1 = 6.20 m/s

$v_{2_{i}}$ : is the initial velocity of the ball 2 = 0 (it is at rest)

$v_{1_{f}}$ : is the final velocity of the ball 1 =?

$v_{2_{f}}$ : is the initial velocity of the ball 2 =?

$m_{1}v_{1_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

$v_{1_{f}} = v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}}$ (1)

Now, by conservation of kinetic energy (since they collide elastically):

$\frac{1}{2}m_{1}v_{1_{i}}^{2} = \frac{1}{2}m_{1}v_{1_{f}}^{2} + \frac{1}{2}m_{2}v_{2_{f}}^{2}$

$m_{1}v_{1_{i}}^{2} = m_{1}v_{1_{f}}^{2} + m_{2}v_{2_{f}}^{2}$ (2)

By entering equation (1) into (2) we have:

$m_{1}v_{1_{i}}^{2} = m_{1}(v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}})^{2} + m_{2}v_{2_{f}}^{2}$

$0.1 kg*(6.20 m/s)^{2} = 0.1 kg*(6.2 m/s - \frac{0.3 kg*v_{2_{f}}}{0.1 kg})^{2} + 0.3 kg(v_{2_{f}})^{2}$

By solving the above equation for $v_{2_{f}}$ :

$v_{2_{f}} = 3.1 m/s$

Now, $v_{1_{f}}$ can be calculated with equation (1):

$v_{1_{f}} = 6.20 m/s - \frac{0.3 kg*3.1 m/s}{0.1 kg} = -3.1 m/s$

The minus sign of $v_{1_{f}}$ means that the ball 1 (100g) is moving in the negative x-direction after the collision.

Therefore, the magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

I hope it helps you!

5 0

3 years ago