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2 years ago

"what is the period of one vibration of this tone?"

1 answer:

7 0

The full question is:
On a keyboard, you strike middle C, whose frequency is 256 Hz. What is the period of one vibration of this tone?
The period of a vibration is the time it takes for the particle to make one full oscillation. Frequency is by definition number of full oscillations per unit of time.
When the frequency is expressed in Hz that unit of time is one second.
So there is the following relation between frequency and period:
$T=\frac{1}{f}$
When we plug in the numbers we get:
$T=\frac{1}{256}=0.0039$s$

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A colloidal liquid in a has is called

S_A_V [24]

Aerosol is the answer

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3 years ago

use what you have learned so far about heat transfer to explain how hot rocks can be used to make steam?

Anna [14]

If you throw water on anything hot enough, the liquid water turns to steam. With this method, you can create an energy source.

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3 years ago

Canadian Olivia Oliver broke the Guinness World Record for the fastest spin on ice skates on January 2015 with an angular veloci

Shalnov [3]

Answer:

Explanation:

initial angular velocity ω' = 130 rpm

final angular velocity ω = 342 rpm

recall that angular momentum = ωI

where I is the moment of inertia.

for the initial spinning condition, we take moment of inertial = I'

for final spinning condition, we take moment of inertia = I

initial angular momentum = ω'I' = 130 I'

final angular momentum = ωI = 342 I

according to conservation of angular momentum, initial angular momentum must be equal to the final angular momentum, therefore

342 I = 130 I'

ratio of initial moment of inertia to final moment of inertia = I'/I

==> I'/I = 342/130 ≅ <em>2.63 : 1</em>

b) to achieve a final angular velocity of of 375 rpm, her initial velocity will have to be

2.63 = 375/ω'

ω' = 375/2.63 = <em>143.13 rpm</em>

6 0

2 years ago

A merry-go-round with a a radius of R = 1.63 m and moment of inertia I = 196 kg-m2 is spinning with an initial angular speed of

Ulleksa [173]

Answer:

299.88 kgm²/s

499.758 kgm²/s

Explanation:

R = Radius of merry-go-round = 1.63 m

I = Moment of inertia = 196 kgm²

$\omega_i$ = Initial angular velocity = 1.53 rad/s

m = Mass of person = 73 kg

v = Velocity = 4.2 m/s

Initial angular momentum is given by

$L=I\omega_i\\\Rightarrow L=196\times 1.53\\\Rightarrow L=299.88\ kgm^2/s$

The initial angular momentum of the merry-go-round is 299.88 kgm²/s

Angular momentum is given by

$L=mvR\\\Rightarrow L=73\times 4.2\times 1.63\\\Rightarrow L=499.758\ kgm^2/s$

The angular momentum of the person 2 meters before she jumps on the merry-go-round is 499.758 kgm²/s

5 0

3 years ago

An object with a density of 250 kg/m3 floats on water. what portion of the object is submerged?

lbvjy [14]

The portion of the object submerged in water is determined as 0.25.

<h3>Fraction of the object submerged in water</h3>

The fraction of the object submerged in water is calculated as follows;

S.G = (density of object) / (density of water)

where;

S.G is specific gravity

S.G = (250 kgm⁻³) / (1000 kgm⁻³)

S.G = 0.25 = ¹/₄

Thus, the portion of the object submerged in water is determined as 0.25.

Learn more about density here: brainly.com/question/1354972

#SPJ1

6 0

1 year ago