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saveliy_v [14]
3 years ago
10

One railroad car rolls towards another. Describe the properties of the first car that determine its momentum and how it would im

part an impulse to the second car.
Physics
1 answer:
notka56 [123]3 years ago
5 0

Answer:

Explanation:

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied

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The friends take a few minutes to discuss their ideas. Which of Amy's comments with respect to a charged particle moving in a ma
Marizza181 [45]

Answer:

The statement "if the magnetic force is always perpendicular to the velocity, the path of the particle is a straight line" is false.

Explanation:

The equation for the magnetic force on a charge q moving at velocity v on a magnetic field B is given by the (vectorial) Lorentz Force Law F=qv\times B

From it we can clearly see that the <em>magnitude of the magnetic force </em>exerted on the particle is <em>proportional to the magnitude of the charge q and to the speed v of the particle</em>, and that it is also <em>perpendicular to the particle's velocity</em>. This means that at each instant it moves perpendicularly to the force, so <em>the work done by the magnetic force on the particle is zero</em>.

The statement "if the magnetic force is always perpendicular to the velocity, the path of the particle is a straight line" is false not only for this but for any force, a force always perpendicular to a velocity will curve the trajectory.

7 0
3 years ago
Si tenemos tres cubos del mismo tamaño (hierro, madera e icopor). ¿Qué diferencias puede encontrar entre ellos?
cricket20 [7]

Answer: La diferencia es el peso (o la masa), siendo que el cubo de hierro es el mas pesado, después viene el de madera y después el de icopor.

Explanation:

Ok, los 3 cubos tienen el mismo tamaño, lo que implica que tienen el mismo volumen.

Ahora es útil recordar la relación:

Densidad = masa/volumen.

Masa = densidad*volumen.

Nosotros sabemos que la densidad del hierro es mas grande que la de la madera, y la densidad de la madera es mas grande que la del icopor.

Entonces, por la relación anterior, y sabiendo que todos los cubos tienen el mismo volumen, podemos reconocer que el cubo de hierro tiene mayor masa, después viene el de madera y después viene el de icopor.

Y sabiendo que:

masa*gravedad = peso

podemos saber que el cubo mas pesado es el de hierro, después el de madera y después el de icopor.

Además de esta diferencia, también hay otras que no dependen tanto del tamaño del objeto, como pueden ser las capacidades caloríficas, el como reaccionan a campos eléctricos y cosas así que son triviales, pues son diferentes para casi todos los materiales.

5 0
3 years ago
Someone places a chocolate bar near a working radar set that is used to locate ships and airplanes. which best describes what is
hjlf
The chocolate softens and melts as it absorbs the microwaves, and it makes a mess.
7 0
3 years ago
Read 2 more answers
Draw a stationary wave and show the position of node and antinode​
Aneli [31]

I hope that the attachment helps you..

3 0
2 years ago
A spring with spring constant 11.5 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to res
Natalija [7]

Answer:

The time constant is \tau = 17.5 \ s    

Explanation:

From the question we are told that

   The spring constant is  k = 11.5 \  N/m

   The mass  of the ball is  m_b  = 490 \ g  = 0.49 \ kg

   The amplitude of the  oscillation t the beginning is x =  6.70 cm = 0.067 \  m

    The amplitude after time t is  x_t = 2.20 cm = 0.022 \  m

    The number of oscillation is N  = 30

Generally the time taken to attain the second amplitude is mathematically represented as

       t  = N  *  T                                            Here  T is the period of oscillation

         t = N * 2\pi \sqrt{\frac{m}{k} }

=>     t = 30 * 2 * 3.142 *  \sqrt{\frac{ 0.490}{11.5} }

=>     t = 38.88 \  s

Generally the amplitude at time t is mathematically represented as

         x(t) = x e^{-\frac{at}{2m} }

Here a is the damping  constant so

 at  t = 38.88 \  s ,  x_t = 2.20 cm = 0.022 \  m

So  

     0.022 = 0.067 e^{-\frac{a * 38.88}{2 * 0.490} }

=>  0.3284 = e^{-\frac{a * 38.88}{2 * 0.490} }

taking natural log of both sides

=>  ln(0.3284 ) = -\frac{a * 38.88}{2 * 0.490} }    

=>   a = 0.028

Generally the time constant is mathematically represented as

    \tau = \frac{m}{a}      

=> \tau = \frac{0.490}{  0.028}    

=> \tau = 17.5 \ s    

4 0
3 years ago
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