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2 years ago

Now that you know how forces affect the motions of objects, you can use the Tracker video analysis tool to create dynamic models

Physics

1 answer:

rodikova [14]2 years ago

4 0

Answer:

Now that you know how forces affect the motions of objects, you can use the Tracker video analysis tool to create dynamic models

for a wide range of physical situations.

Tracker enables you to create two different types of mathematical models: analytical and dynamic. An analytical model enables you

to enter mathematical expressions for x and y positions as a function of time. That's sometimes useful, but from a physics

perspective, a dynamic model is much more flexible and powerful.

A dynamic model enables you to set the initial conditions for a particular system (Initial positions and velocities); then you can

mathematically define any forces acting on that system. Once those are set up, the model acts like an object in space, responding to

the forces you've imposed on it. It can continue moving forever, if that's what the forces would do to an object in real life. By visually

matching a marker for your model to the real motion on the video, you can define and refine a mathematical model for a

You might be interested in

A car has a mass of 1000 kg. What is the acceleration produced by a force of 2000 N?

EastWind [94]

F=ma
a=F/m
a=2000/1000
a=2 m/s^2

6 0

2 years ago

A 5.00 kg rock whose density is 4300 kg/m3 is suspended by a string such that half of the rock's volume is under water. You may

12345 [234]

Answer:

The tension on the string is $T = 43.302 \ N$

Explanation:

From the question we are told that

The mass of the rock is $m_r = 5.00 \ kg = 5000 \ g$

The density of the rock is $\rho = 4300 \ kg/m^3 = 4.3 g/dm^3$

Generally the volume of the rock is mathematically evaluated as

$V = \frac{m_r}{\rho}$

substituting values

$V = \frac{5000}{4.3}$

$V = 1162.7 \ dm^3$

The volume of the rock immersed in water is

$V_w = \frac{V}{2}$

substituting values

$V_w = \frac{1162.7 }{2}$

$V_w = 581.4 \ dm^3$

mass of water been displaced by the this volume is

$m_w = V_w$ According to Archimedes principle

=> $m_w = 581.4 \ g$

$m_w = 0.5814 \ kg$

The weight of the water displace is

$W _w = m_w * g$

$W _w = 0.5814 * 9.8$

$W _w = 5.698 \ N$

The actual weight of the rock is

$W_r = m_r * g$

$W_r = 5.0 * 9.8$

$W_r = 49.0 \ N$

The tension on the string is

$T = W_r - W_w$

substituting values

$T = 49.0 - 5.698$

$T = 43.302 \ N$

4 0

3 years ago

A Styrofoam ball has just been shot straight up. Air resistance is not negligible (free body diagram)

timofeeve [1]

In a free body diagram for an object projected upwards;

the acceleration due to gravity on the object is always directed downwards.

the velocity of the object is always in the direction of the object's motion.

An object projected upwards is subjected to influence of acceleration due to gravity.

As the object accelerates upwards, its velocity decreases until the object reaches maximum height where its velocity becomes zero and as the object descends its velocity increases, which eventually becomes maximum before the object hits the ground.

To construct a free body diagram for this motion, we consider the following;

the acceleration due to gravity on the object is always directed downwards

the velocity of the object is always in the direction of the object's motion.

<u>For instance:</u>

upward motion for velocity ↑ downward motion for velocity ↓

↑ ↓

acceleration due to gravity ↓

↓

Learn more here: brainly.com/question/13235430

5 0

3 years ago

A 2.0-kg object moving with a velocity of 5.0 m/s in the positive x direction strikes and sticks to a 3.0-kg object moving with

Andrej [43]

Answer:

5.4 J.

Explanation:

Given,

mass of the object, m = 2 Kg

initial speed, u = 5 m/s

mass of another object,m' = 3 kg

initial speed of another orbit,u' = 2 m/s

KE lost after collusion = ?

Final velocity of the system

Using conservation of momentum

m u + m'u' = (m + m') V

2 x 5 + 3 x 2 = ( 2 + 3 )V

16 = 5 V

V = 3.2 m/s

Initial KE = $\dfrac{1}{2}mu^2 + \dfrac{1}{2}m'u'^2$

= $\dfrac{1}{2}\times 2\times 5^2 + \dfrac{1}{2}\times 3 \times 2^2$

= 31 J

Final KE = $\dfrac{1}{2} (m+m')V^2 = \dfrac{1}{2}\times 5 \times 3.2^2 = 25.6 J$

Loss in KE = 31 J - 25.6 J = 5.4 J.

4 0

3 years ago

A block of mass m=9.0 kg and speed V and is behind a block of mass M= 27 kg and speed of .50 m/s. The surface is frictionless, a

sammy [17]

Answer:

2.06 m/s

Explanation:

From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.

Momentum before collision

Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5

Momentum after collision

The momentum after collision will be given by (9+27)*0.9=32.4

Relating the two then 9v+13.5=32.4

9v=18.5

V=2.055555555555555555555555555555555555555 m/s

Rounded off, v is approximately 2.06 m/s

5 0

3 years ago