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Fudgin [204]
3 years ago
15

Two blocks on a frictionless horizontal surface are on a collision course. one block with mass 0.6 kg moves at 0.8 m/s to the ri

ght and collides with a 1.2 kg mass at rest and the two masses bounce off each other elastically. find the final velocities of the two masses after the collision.
Physics
1 answer:
Elanso [62]3 years ago
8 0

First we can say that since there is no external force on this system so momentum is always conserved.

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

0.6*0.8 + 1.2*0 = 0.6*v_{1f} + 1.2*v_{2f}

0.48= 0.6*v_{1f} + 1.2*v_{2f}

0.8  = v_{1f} + 2v_{2f}

now by the condition of elastic collision

v_{2f} - v_{1f} = 0.8 - 0[\tex]now add two equations[tex]3*v_{2f} = 1.6

v_{2f} = 0.533 m/s

also from above equation we have

v_{1f} = -0.267 m/s

So ball of mass 0.6 kg will rebound back with speed 0.267 m/s and ball of mass 1.2 kg will go forwards with speed 0.533 m/s.

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Small, icy bodies that have highly eccentric orbits and can be found in the Oort cloud or the Kuiper belt are called ____.
irina1246 [14]

Answer:

Small, icy bodies that have highly eccentric orbits and can be found in the Oort cloud or the Kuiper belt are called COMETS.

4 0
3 years ago
Si un cuerpo adquiere una carga de -0,02 C, ha ganado o ha perdido electrones? Cuantos?
sammy [17]

Question :

If a body acquires a charge of -0.02 C, has it gained or lost electrons? Many?

Solution :

We know, charge gained is shown by negative sign.

Since, charged acquired is given as -0.02 C .

Therefore, it is body has gained electrons.

Now, number of electrons is given by :

n = \dfrac{net\ charge}{charge \  on \ one \  electron}\\\\n = \dfrac{-0.02}{-1.60 \times 10^{-19}}\\\\n = 1.25\times 10^{17}

Hence, this is the required solution.

7 0
3 years ago
A block of mass
Gennadij [26K]

(a) The work done by the applied force is 26.65 J.

(b) The work done by the normal force exerted by the table is 0.

(c) The work done by the force of gravity is 0.

(d) The work done by the net force on the block is 26.65 J.

<h3>Work done by the applied force</h3>

W = Fdcosθ

W = 14 x 2.1 x cos25

W = 26.65 J

<h3>Work done by the normal force</h3>

W = Fₙd

W = mg cosθ x d

W = (2.5 x 9.8) x cos(90) x 2.1

W = 0 J

<h3>Work done force of gravity</h3>

The work done by force of gravity is also zero, since the weight is at 90⁰ to the displacement.

<h3> Work done by the net force on the block</h3>

∑W = 0 + 26.65 J = 26.65 J

Thus, the work done by the applied force is 26.65 J.

The work done by the normal force exerted by the table is 0.

The work done by the force of gravity is 0.

The work done by the net force on the block is 26.65 J.

Learn more about work done here: brainly.com/question/8119756

#SPJ1

6 0
2 years ago
when the heat is added to the material, which factors are important in order to determine how much the material will rise in tem
posledela

The heat Q transferred to cause a temperature change depends on the magnitude of the temperature change, the mass of the system, and the substance and phase involved.

Explanation:

https://courses.lumenlearning.com/physics/chapter/14-2-temperature-change-and-heat-capacity/

5 0
3 years ago
The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s. (a) What
Neko [114]

Complete Question

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.

(a) What is its angular acceleration in revolutions per minute-squared

(b) How many revolutions does the engine make during this 20 s interval?

rev

Answer:

a

 \alpha = 6261 \  rev/minutes^2

b

 \theta  = 613 \ revolutions

Explanation:

From the question we are told that

   The initial  angular speed is w_i =  1120 \ rev/minutes

    The angular speed after t = 13.8 s = \frac{13.8}{60 }  = 0.23 \ minutes  is w_f = 2560 \ rev/minutes

    The time for revolution considered is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  

 Generally the angular acceleration is mathematically represented as

         \alpha = \frac{w_f - w_i }{t}

=>      \alpha = \frac{2560  - 1120 }{0.23}  

=>      \alpha = 6261 \  rev/minutes^2

Generally the number of revolution made is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  is mathematically represented as

           \theta  =  \frac{1}{2}  * (w_i + w_f)*  t

=>      \theta  =  \frac{1}{2}  * (1120+ 2560 )*  0.333

=>      \theta  = 613 \ revolutions

5 0
3 years ago
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