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alex41 [277]
3 years ago
10

Two balls, each with a mass of 0.5 kg, collide on a pool table. Is the law of conservation of momentum satisfied in the collisio

n? Explain why or why not
Physics
2 answers:
Dmitriy789 [7]3 years ago
7 0

During the collision between two balls on the pool table there is no external force along the line of collision between them

Since there is no external force on it so here we can say

F = 0 = \frac{\Delta P}{\Delta t}

here we have

\Delta P = 0

so we can say

P_i = P_f

since there is no external force so we can say during the collision the momentum of two balls will remain conserved

aleksklad [387]3 years ago
7 0
<h3>Sample Response: <em>Yes, the law of conservation of momentum is satisfied. The total momentum before the collision is 1.5 kg • m/s and the total momentum after the collision is 1.5 kg • m/s. The momentum before and after the collision is the same.</em></h3><h3><em /></h3>

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A polarized light that has an intensity I0 = 60.0 W/m² is incident on three polarizing disks whose planes are parallel and cente
nikitadnepr [17]

Answer:

The transmitted intensity through all polarizers is I_3 =41.31 W/m^2

Explanation:

 According to Malu's law the intensity of a polarized light having an initial intensity I_0 is mathematically represented as

               I = I_0cos^2 \theta

Now  considering the polarizer(The polarizing disk) the equation above becomes

          I = I_0 (cos^2 \theta)^n

Where n is the number of polarizers

       Substituting  60.0W/m^2 for the initial intensity 3 for the n and 20° for the angle of rotation

           I_3 = 60 (cos^220)^3

               =41.31 W/m^2

             

     

                         

6 0
3 years ago
in a closed system three objects have the following momentum: 11 kg* m/s, -65 kg*m/s and -100 kg m/s. the objects collide and mo
guajiro [1.7K]

Explanation:

The momentum of the three objects are as follow :

11 kg-m/s, -65 kg-m/s and -100 kg-m/s

Before collision, the momentum of the system is :

P_i=11+(-65)+(-100)\\\\P_i=-154\ kg-m/s

After collison, they move together. It means it is a case of inelastic collision. In this type of collision, the momentum of the system remains conserved.

It would mean that, after collision, momentum of the system is equal to the initial momentum.

Hence, final momentum = -154 kg-m/s.

4 0
3 years ago
With the piston head locked in place, will the volume of the gas increase, decrease, or stay the same whenthe piston is placed a
Nuetrik [128]

Answer:

For real gas the volume of a given mass of gas will increase with increase in temperature.

Explanation:

With the piston head locked in place and place above the fire,the volume of the gas will increase,because the volume of a given mass of gas increases with increase temperature.

6 0
3 years ago
Why does the candle light has no shadow when light fall on it?​
andrezito [222]
Shadows are the absence of light, they are created when an object blocks light. In other words, shadows are the product of light particles, known as photons. These particles “bounce off” of the object without reaching the other side. Therefore light by itself will not form a shadow.
6 0
3 years ago
A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the cap
viktelen [127]

Answer:

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

V_r + V_L + V_c = V_{net}

now we will have

iR + L\frac{di}{dt} + \frac{q}{C} = 12 V

now we have

1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12

So we will have

q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}

at t = 0 we have

q = 0

0 = 3\times 10^{-6} + c_1  + c_2

also we know that

at t = 0 i = 0

0 = -4000 c_1 - 1000c_2

c_2 = -4c_1

c_1 = 1 \times 10^{-6}

c_2 = -4 \times 10^{-6}

so we have

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

5 0
3 years ago
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