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3 years ago

What is 174HZ and its pitch

1 answer:

8 0

The lowest recognized solfeggio frequency is 174 Hz, also known as the healing frequency. It has the greatest impact on the body and physical health, rather than spiritual or mental health. People mainly experience pain relief from this frequency, although it also reduces stress.

Answer: 174 Removes pain.

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Thomson's cathode ray beam bent towards the positively charged plate. With this information, he was able to conclude

andre [41]

I believe the correct answer from the choices listed above is the second option. With this information, he was able to conclude that the beam was negatively charged. His cathode ray tube experiments led to a very important scientific discovery, the electron. Hope this answers the question.

8 0

3 years ago

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Given the Earth's density as 5.5 g/cm3 and the Moon's density as 3.34 g/cm3, determine the Roche limit for the Moon orbiting the

Assoli18 [71]

Answer:

The Roche limit for the Moon orbiting the Earth is 2.86 times radius of Earth

Explanation:

The nearest distance between the planet and its satellite at where the planets gravitational pull does not torn apart the planets satellite is known as Roche limit.

The relation to determine Roche limit is:

$Roche\ limit=2.423\times R_{P}\times\sqrt[3]{\frac{D_{P} }{D_{M} } }$ ....(1)

Here $R_{P}$ is radius of planet and $D_{P}\ and\ D_{M}$ are density of planet and moon respectively.

According to the problem,

Density of Earth, $D_{P}$ = 5.5 g/cm³

Density of Moon, $D_{M}$ = 3.34 g/cm³

Consider $R_{E}$ be the radius of the Earth.

Substitute the suitable values in the equation (1).

$Roche\ limit=2.423\times R_{E}\times\sqrt[3]{\frac{5.5 }{3.34 } }$

$Roche\ limit= 2.86R_{P}$

8 0

3 years ago

(b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational

Oksanka [162]

Answer:

Explanation:

a) using the energy conservation equation

mgh = 0.5mv^2 + 0.5Iω^2

I(moment of inertia) (basket ball) = (2/3)mr^2

mgh = 0.5mv^2 + 0.5( 2/3mr^2) ( v^2/r^2)

gh = 1/2v^2 + 1/3v^2

gh = v^2( 5/6)

v = $\sqrt{\frac{6gh}{5} }$

putting the values we get

$6.6 ^{2} = \frac{6\times9.8h}{5}$

solving for h( height)

h = 3.704 m apprx

b) velocity of solid cylinder

mgh = 0.5mv^2 + 0.5( mr^2/2)( v^2/r^2) where ( I ofcylinder = mr^2/2)

g*h = 1/2v^2 + 1/4v^2

g*h = 3/4v^2

putting the value of h and g we get

v= = 6.957 m/s apprx

5 0

4 years ago

A. 7cm B. 24cm C. 18cm D. 63cm

Alex17521 [72]

Answer:

A. 7cm

I hope it helps...

6 0

3 years ago

A ball rolls off the edge of a horizontal roof at 10 m/s. 2 seconds later, the speed of the ball will be? a. 10 m/s b. 19.6m/s c

disa [49]

The ball rolls horizontally thus it is not affected by the gravitational acceleration. We assume that velocity is constant since we are not given acceleration. Therefore, acceleration is zero. The speed of the ball after 2 seconds will still be the same given we have a constant velocity system. Option A is the answer.

6 0

4 years ago

Read 2 more answers