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IRINA_888 [86]
2 years ago
12

A cue ball, moving with 9.0 N·s of momentum strikes the nine-ball at rest. The nine-ball moves off with 2.0 N·s in the original

direction of the cue ball and 2.0 N·s perpendicular to that direction. What is the momentum of the cue ball after the collision?
Physics
1 answer:
lisov135 [29]2 years ago
7 0

Answer:

P = 7.28 N.s

Explanation:

given,

initial momentum of cue ball in x- direction,P₁ = 9 N.s

momentum of nine ball in  x-  direction, P₂ = 2 N.s

momentum in perpendicular direction i.e. y - direction,P'₂ = 2 N.s

momentum of the cue after collision = ?

using conservation of momentum

in x- direction

P₁ + p = x  + P₂

p is the initial momentum of the nine balls which is equal to zero.

9 + 0  = x  + 2

x = 7 N.s

momentum in x-direction.

equating along y-direction

P'₁ + p = y + P'₂

0 + 0 = y + 2

y = -2 N.s

the momentum of the cue ball after collision is equal to resultant of the momentum .

P = \sqrt{x^2+y^2}

P = \sqrt{7^2+(-2)^2}

      P = 7.28 N.s

the momentum of the cue ball after collision is equal to P = 7.28 N.s

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A light with a second-order bright band forms a diffraction angle of 30. 0°. The diffraction grating has 250. 0 lines per mm. Wh
Luden [163]

The distance between two successive troughs or crests is known as the wavelength. The wavelength of the light will be 1000 nm.

How do you define wavelength?

The distance between two successive troughs or crests is known as the wavelength. The peak of the wave is the highest point, while the trough is the lowest.

The wavelength is also defined as the distance between two locations in a wave that have the same oscillation phase.

Diffraction angle= 30⁰

Diffraction grating per mm= 250

wavelength = ?

Mathematically the equation of bright band is given by

\rm \lambda= \frac{sin\theta}{nN}

\rm \lambda= \frac{sin23^0}{250\times 2}

\rm \lambda= 0.000001 m

\rm \lambda= 1000 nm

Hence the wavelength of the light will be 1000 nm.

To learn more about the wavelength refer to the link;

brainly.com/question/7143261

8 0
2 years ago
Read 2 more answers
The Balmer series in hydrogen atom produces only infrared emission. O True O False
lorasvet [3.4K]

Answer:

False

Explanation:

As we know that, the Balmer series gives the n values as,

n_{i}=2.

[tex]n_{f}=3,4,5,.....\infty.

Now the value of wavelength can be calculated as,

\frac{1}{\lambda}=R(\frac{1}{n_{i} }-\frac{1}{n_{f} } )z^{2}.

Here, R=109677 cm^{-1}.

And n_{f}=3.

Now,

\frac{1}{\lambda}=109677 cm^{-1}(\frac{1}{2}-\frac{1}{3}).

Therefore,

\lambda=\frac{6}{109677} cm\\\lambda=547\times 10^{-9} m\\ \lambda=547 nm

Therefore, the wavelength of Balmer series lies in visible region which is 547 nm.

8 0
3 years ago
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A two-stage rocket moves in space at a constant velocity of +4300 m/s. The two stages are then separated by a small explosive ch
ololo11 [35]

Answer:

 v_{1f} = +3,394 103 m / s

Explanation:

We will solve this problem with the concept of the moment. Let's start by defining the system that is formed by the complete rocket before and after the explosions, bone with the two stages, for this system the moment is conserved.

The data they give is the mass of the first stage m1 = 2100 kg, the mass of the second stage m2 = 1160 kg and its final velocity v2f = +5940 m / s and the speed of the rocket before the explosion vo = +4300 m / s

The moment before the explosion

      p₀ = (m₁ + m₂) v₀

After the explosion

      pf = m₁ v_{1f} + m₂ v_{2f}

     p₀ = [texpv_{f}[/tex]

     (m₁ + m₂) v₀ = m₁ v_{1f} + m₂ v_{2f}

Let's calculate the final speed (v1f) of the first stage

     v_{1f} = ((m₁ + m₂) v₀ - m₂ v_{2f}) / m₁

     

     v_{1f} = ((2100 +1160) 4300 - 1160 5940) / 2100

     v_{1f} = (14,018 10 6 - 6,890 106) / 2100

     v_{1f} = 7,128 106/2100

     v_{1f} = +3,394 103 m / s

come the same direction of the final stage, but more slowly

4 0
3 years ago
The gage pressure in a liquid at a depth of 3 m is read to be 39 kPa. Determine the gage pressure in the same liquid at a depth
ioda

Answer: 117 kPa

Explanation:

For the liquid at depth 3 m, the gauge pressure is equal to = P₁=39 kPa

For the liquid at depth 9m, the gauge pressure is equal to= P₂

Now we are given the condition that the liquid is same. That must imply that the density must be same throughout the depth.

So, For finding gauge pressure we have formula P= ρ * g * h

Also gravity also remains same for both liquids

So taking ratio of their respective pressures we have

\frac{P_{1} }{\\P_2}= \frac{density * g * h_1}{density * g * h_2}

So \frac{39}{P_2}= \frac{3}{9}

Or P₂= 39 * 3 = 117 kPa

5 0
3 years ago
Please. Physics is so difficult.
Softa [21]

Answer:

0.010 m

Explanation:

So the equation for a pendulum period is: y=2\pi\sqrt{\frac{L}{g}} where L is the length of the pendulum. In this case I'll use the approximation of pi as 3.14, and g=9.8 m\s. So given that it oscillates once every 1.99 seconds. you have the equation:

1.99 s = 2(3.14)\sqrt{\frac{L}{9.8 m\backslash s^2}}\\

Evaluate the multiplication in front

1.99 s = 6.28\sqrt{\frac{L}{9.8m\backslash s^2}

Divide both sides by 6.28

0.317 s= \sqrt{\frac{L}{9.8 m\backslash s^2}}

Square both sides

0.100 s^2= \frac{L}{9.8 m\backslash s^2}

Multiply both sides by m/s^2  (the s^2 will cancel out)

0.984 m = L

Now now let's find the length when it's two seconds

2.00 s = 6.28\sqrt{\frac{L}{9.8m\backslash s^2}}

Divide both sides by 6.28

0.318 s = \sqrt{\frac{L}{9.8 m\backslash s^2}

Square both sides

0.101 s^2 = \frac{L}{9.8 m\backslash s^2}

Multiply both sides by 9.8 m/s^2 (s^2 will cancel out)

0.994 m = L

So to find the difference you simply subtract

0.984 - 0.994 = 0.010 m

4 0
1 year ago
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