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kodGreya [7K]
3 years ago
5

A 2kg object travels to the right at 5.5 m/s after colliding with a 3kg object, initially at rest, what are the speeds of each o

bject after the collision l?

Physics
1 answer:
Snowcat [4.5K]3 years ago
3 0

Answer:

Velocity of Object with 2 kg= 3.390 m/s

Velocity of Object with 3 kg= 3.404 m/s

Explanation:

From the picture, it can be seen that object B is initially at rest while object A is travelling at a speed of 5m/s. After the collision, Object A moves at an angle of 65 degrees while object B moves at an angle of 37 degrees.

We also know that momentum of a closed system is conserved.

Initial momentum along the x-axis = 2*5.5 = 11

Initial momentum along y-axis = 0

Final momentum along x-axis= a*Cos(65)*2 +b*Cos(37) *3= 11 (a is the velocity of object A of 2 kg after collision where as b is the velocity of object B of 3 kg after collision. velocity is multiplied by cosines of the angle from x axis to give the horizontal component of the velocities).

Final momentum along y-axis = a*Sin(65)*2 - b*Sin(37)*3 =0 (We can see that vertical components of velocity are opposite in direction to each other)

Solve both the equations simultaneously for a and b.

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A football punter wants to kick the ball so that it is in the air for 4.5 s and lands 50 m from where it was kicked. Assume that
irakobra [83]

Answer:

(a) The angle of projection is 63 degree.

(b) The velocity of projection is 24.5 m/s.

Explanation:

Height, h = 1 m

horizontal distance, d = 50 m

time, t = 4.5 s

Let the initial velocity is u and the angle is A.

(a) Horizontal distance = horizontal velocity x time

50 = u cos A x 4.5

u cos A = 11.1 .....(1)

Use second equation of motion in vertical direction

h = u t + 0.5 gt^2\\\\- 1 = u sin A \times 4.5 - 0.5 \times 9.8\times 4.5^2\\\\u sin A = 21.8 ..... (2)

Divide (2) by (1)

tan A = 1.97

A = 63 degree

(b) Substitute the value of A in equation (2)

u x sin 63 = 21.8

u = 24.5 m/s

7 0
3 years ago
I The Circuit shows for Tamps Connected by Tovde supply
luda_lava [24]
.........,....,,,???........
4 0
2 years ago
Which strategy should you use if your research question is too broad for the scope of your project? (1 O narrow the focus of res
Nesterboy [21]

Answer:

"Narrow the focus of research question"

Explanation:

O Narrow the focus of research question

    This is good! You can still use your question, but focus in on something so you have a proper research project.

O Add another research question

    Would adding another question to an already broad question help? No.

O Use the very first source you find for your project

    If your question is too broad, you should not use whatever you see first as it may be incorrect or does not answer the question

O Change the scope of your project​    

    You could, but if you have a set scope for your project (a) you might not be able to change it (b) you don't need to restart

Have a nice day!

    I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

3 0
2 years ago
If 0.250 L of a gas weighs 0.308 g under normal conditions of pressure and temperature, what is its molecular weight?
Phantasy [73]

Answer:

=28

Explanation:

.25/22=0,11 mol

0.308/0,11=28

8 0
3 years ago
A diffraction grating with 1000 lines per mm is used in a spectrometer to measure the wavelengths of light emitted from a gas di
frez [133]

Answer:

= 9.8°

Explanation:

Width of one slit (a₁ ) = 1 / 1000 mm=0.001 mm = 10⁻⁶ m.

width of one slit in case 2 (a₂ ) = 1/500 =2 x 10⁻⁶ m

angular position of fringe, Sinθ  = n λ /a

n is order of fringe , λ is wave length of light and a  is slit aperture

So Sinθ  ∝ 1 / a

Sin θ₁ /Sin θ₂ = a₂/a₁ ;

Sin20°/sinθ₂ = 2 / 1

sinθ₂ = Sin 20° / 2 = .342/2 = .171

θ₂ = 9.8 °

4 0
3 years ago
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