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3 years ago

A ball rolls off a table and it traveling with a horizontal velocity of 2 m/s and 1 point

Physics

2 answers:

morpeh [17]3 years ago

4 0

Answer:

The velocity when the ball hits the ground is obtained using v2. 2 = v1. 2 + 2 g Dy with v1=0 and Dy=h. Thus solving for v2 yields 17.1 m/s v2 = 2 g h =.

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astra-53 [7]3 years ago

4 0

Answer:

I e jofh ndoj eohe nc n very nic e

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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

A car travels a distance of 140 km at 70.0 km/hr. It then travels an additional distance of 60.0 km at 40.0 km/hr. The average s

Natali5045456 [20]

Answer:

57.1 km/hr

Explanation:

To find the average speed you take the total distance divided by the total elapsed time.

So, the total distance is 140 + 60 = 200

the total elapsed time is found by taking 140/70=2 and 60/40=1.5

2+1.5=3.5

The plug the numbers into the equation,

200/3.5=57.1

5 0

3 years ago

Vector A has a magnitude 5 m, vector B has a magnitude 7 m. Which of the following cannot be the magnitude of the vector A + B?

bulgar [2K]

I think it should be 7m

7 0

4 years ago

consider the free-body diagram. if you want the box to move, the force applied while dragging must be greater than the

NeTakaya

Answer:

Force of static friction between the two surfaces

Explanation:

When two surfaces come into contact, they exert a force that resist the sliding of the two surfaces. This force is called static friction.

This force is given by the relation

$F_{s}=\mu_{s}\eta$

Where,

μ - coefficient of static friction

η - normal force acting on the body

When a force acts on a body placed on a rough surface, it doesn't do any work if the applied force was less than the force of static friction.

So, in order to move the body, the applied force should be greater than the force of static friction.

6 0

3 years ago

How do elements change as you move down a column in a periodic table? a.the atomic radius decreases b.the number of protons decr

MariettaO [177]

As you move down a column :
* the atomic number increases
* the number of protims increases
* the atomic mass increases
* the electronegativity decreases

4 0

4 years ago

Read 2 more answers