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SSSSS [86.1K]
2 years ago
8

Describe briefly how you would determine the density an irregular object (stone)

Physics
1 answer:
4vir4ik [10]2 years ago
6 0

Answer:

by measuring calender we can find

Explanation:

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A boy throws a ball and accidentally breaks a window. The momentum of the ball and all the pieces of glass taken together after
Aleks [24]
When a boy throws a ball and accidentally breaks a window, the momentum of the ball and all the pieces of glass taken together after the collision is THE SAME as the momentum of the ball before the collision

hope this helps
6 0
3 years ago
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What do Oxygen and Fluorine do to do Copper?
Evgen [1.6K]
It corrodes the copper by oxydation
4 0
3 years ago
SHOW WORK
Helga [31]

Answer:

Follows are the solution to the given question:

Explanation:

For point a:

T= 2\pi \sqrt{\frac{m}{k}}\\\\k = \frac{4 \pi^2 m}{T^2}\\\\= \frac{4 \times (3.14)^2 \times 3}{2^2}\\\\=29.578 \ \frac{N}{m}\\\\

For Point b:

E=\frac{1}{2} m a^2 w^2\\\\

   =\frac{1}{2} \times m \times a^2 \times \frac{4\pi^2}{T^2}\\\\=\frac{1}{2} \times 3 \times (0.15)^2 \times \frac{4\times 3.14^2}{2^2}\\\\=0.332 \ J

For Point C:

V_{max}= a w

        = (0.15) \times \frac{2\pi}{T}\\\\= (0.15) \times \frac{2\times 3.14}{2}\\\\=0.471 \frac{m}{s}

For point D:

X= a \sin (wt+ \phi)\\\\0.91=0.15 \sin(\frac{2\pi}{T} \times t+\phi)\\\\0.91=0.15 \sin(\frac{2\times 3.14}{2} \times 0.5+\phi)\\\\0.60 = \sin(3.14 \times 0.5+\phi)\\\\0.60 = \sin(1.57+\phi)\\\\1.57 +\phi =\sin^{-1} 60^{\circ}\\\\1.57 +\phi = 36.86^{\circ}\\\\=35.29^{\circ}\\\\So, X=15 \sin(3.14t+35.29^{\circ}) \ cm

5 0
3 years ago
In the picture of the atom above, what subatomic particle does the letter A represent?
ra1l [238]

Answer:

Electron

Explanation:

In the picture, the letter A is pointing to an electron.

4 0
2 years ago
A mass on a spring A oscillates at twice the frequency of the same mass on spring B. Which statement is correct?A.The spring con
Nataliya [291]

Answer:

A.The spring constant for B is one quarter of the spring constant for A.

Explanation:

If spring A oscillates at twice the frequency of spring B, and period is frequency inverted. It means spring B has a period twice of spring A's.

T_B = 2T_A

As T = 2\pi\sqrt{\frac{m}{k}}, and the 2 springs have the same mass

2\pi\sqrt{\frac{m}{k_B}} = 2\pi\sqrt{\frac{m}{k_A}}

\sqrt{k_A} = 2\sqrt{B}

k_A = 4k_B

k_B = k_A/4

So A.The spring constant for B is one quarter of the spring constant for A. is the correct answer.

3 0
3 years ago
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