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3 years ago

PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP

Physics

2 answers:

Alex73 [517]3 years ago

5 0

The bottom thing doesn’t matter. That is not meant to be red, it‘s the bottom of the beaker. The star is at the very bottom of the beaker. it’s just the base of the beaker.

alisha [4.7K]3 years ago

4 0

Answer:

There are 4 liquids in this experiment and red is the least dense of all of them so it should float on top, which it is doing.

The red that you see at the bottom is neither liquid nor is it a part of the experiment.

It is simply the <u>color of the bottom of the container</u> that the experiment was conducted in.

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What is its time interval between the release

azamat

There are several many equations that are available to relate the distance,
speed, and time of a body moving vertically in gravity. Happily, the only one
I can always remember without looking it up happens to be the right one to
use for this question !

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Divide each side
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3 years ago

A disk with mass 1.64 kg and radius 0.61 meters is spinning counter-clockwise with an angular velocity of 17.6 rad/s. A rod of m

Masja [62]

Answer:

The loss in rotational kinetic energy due to the collision is 36.585 J.

Explanation:

Given;

mass of the disk, m₁ = 1.64 kg

radius of the disk, r = 0. 61 m

angular velocity of the disk, ω₁ = 17.6 rad/s

mass of the rod, m₂ = 1.51 kg

length of the rod, L = 1.79 m

angular velocity of the rod, ω₂ = 5.12 rad/s (clock-wise)

let the counter-clockwise be the positive direction

let the clock-wise be the negative direction

The common final velocity of the two systems after the collision is calculated by applying principle of conservation of angular momentum ;

m₁ω₁ + m₂ω₂ = ωf(m₁ + m₂)

where;

ωf is the common final angular velocity

1.64 x 17.6 + 1.51(-5.12) = ωf(1.64 + 1.51)

21.1328 = ωf(3.15)

ωf = 21.1328 / 3.15

ωf = 6.709 rad/s

The moment of inertia of the disk is calculated as follows;

$I_{disk} = \frac{1}{2} mr^2\\\\I_{disk} = \frac{1}{2} (1.64)(0.61)^2\\\\I_{disk} = 0.305 \ kgm^2$

The moment of inertia of the rod about its center is calculated as follows;

$I_{rod} = \frac{1}{12} mL^2\\\\I_{rod} = \frac{1}{12} \times 1.51 \times 1.79^2\\\\I _{rod }= 0.4032\ kgm^2$

The initial rotational kinetic energy of the disk and rod;

$K.E_i = \frac{1}{2} I_{disk}\omega _1 ^2 \ \ + \ \ \frac{1}{2} I_{rod}\omega _2 ^2 \\\\K.E_i= \frac{1}{2} (0.305)(17.6) ^2 \ \ + \ \ \frac{1}{2} (0.4032)(-5.12) ^2\\\\K.E_i = 52.523 \ J$

The final rotational kinetic energy of the disk-rod system is calculated as follows;

$K.E_f = \frac{1}{2} I_{disk}\omega _f ^2 \ \ + \ \ \frac{1}{2} I_{rod}\omega _f ^2\\\\K.E_f = \frac{1}{2} \omega _f ^2(I_{disk} + I_{rod})\\\\K.E_f = \frac{1}{2} (6.709) ^2(0.305+ 0.4032)\\\\K.E_f = 15.938 \ J$