3,89,988 cm/min is the linear velocity
Given,
Diameter of CD = 12 cm
So, Radius of CD = 6 cm
CD is spinning at 10350 rev/min
Firstly , convert rev/min into rad/min
1 rev = 2π radians
10350 rev/min = 10350 × 2π
= 64998 rad/min
Formula used,
where,
is the Linear velocity
is the radius
is the angular velocity
= 6 cm × 64998rad/min
= 3,89,988 cm/min
Thus, linear velocity for any edge point of a 12-cm-diameter CD (compact disc) spinning at 10,350 rev/min is 389988 cm/min.
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Answer:
the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is 
Explanation:
The free-body diagram below shows the interpretation of the question; from the diagram , the wheel that is rolling in a clockwise directio will have two velocities at point P;
- the peripheral velocity that is directed downward
along the y-axis
- the linear velocity
that is directed along the x-axis
Now;
Also,
where 
(angular velocity) = 
∴ the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is
Answer: so when a turbine converts the K.E and the potential of any moving fluid (more likely liquid or gas) to energy. once the proc is started the turbine generato, the fluid such as water, steam, combus gasses, or air pushes s big series of blades that have mounted on a shaft, which then will rotate the shaft that’s conn to the generator
Explanation: hope this helped plz mark brainest
Answer:
Krypton – Mass Number – Neutron Number – Kr 2020-11-21 by Nick Connor Krypton is a chemical element with atomic number 36 which means there are 36 protons and 36 electrons in the atomic structure. The chemical symbol for Krypton is Kr.
Atomic Number: 36
Element: Krypton
Element Category: Noble Gas
Symbol: Kr
Explanation:
Question
Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles travelled in a straight line and some were deflected at different angles.
Which statement best describes what Rutherford concluded from the motion of the particles?
A) Some particles travelled through empty spaces between atoms and some particles were deflected by electrons.
B) Some particles travelled through empty parts of the atom and some particles were deflected by electrons.
C) Some particles travelled through empty spaces between atoms and some particles were deflected by small areas of high-density positive charge in atoms.
D) Some particles travelled through empty parts of the atom and some particles were deflected by small areas of high-density positive charge in atoms.
Answer:
The right answer is C)
Explanation:
In the experiment described above, a piece of gold foil was hit with alpha particles, which have a positive charge. Alpha particles <em>α</em> were used because, if the nucleus was positive, then it would deflect the positive particles. The principles of physics posit that electric charges of the same orientation repel.
So most as expected some of the alpha particles went right through meaning that the gold atoms comprised mostly empty space except the areas that were with a dense population of positive charges. This area became known as the "nucleus".
Due to the presence of the positive charges in the nucleus, some particles had their paths bent at large angles others were deflected backwards.
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