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3 years ago

How do Air molecules enable sound to travel from a radios speaker to your ears?

2 answers:

6 0

The radio frequencies push one air molecule that then bumps into a different air molecule.....which then hits another and another causing a line of crashing molecules that lead inside your ear and hits your ear drum causing it to vibrate which causes the sounds.

madreJ [45]3 years ago

6 0

The molecules bounce and colide and hit like dominoes and vibrate and we haer it

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A rocket's mass decreases as it burns fuel. the equation of motion for a rocket in vertical flight can be obtained from newton's

Talja [164]

The equation of motion for a rocket in vertical flight can be obtained from newton’s second law of motion and is constant-mass system. The equation of motion for a body mass varies with time and mass. When force acts on rocket, the rocket will accelerate in the direction of force. Therefore, force is equal to the change in momentum per change in time. For constant mass, force equals mass times acceleration.

4 0

3 years ago

An arrow is shot vertically upward at a rate of 250ft/s. Use the projectile formula h=−16t2+v0t to determine at what time(s), in

SIZIF [17.4K]

Answer:

The arrow is at a height of 500 feet at time t = 2.35 seconds.

Explanation:

It is given that,

An arrow is shot vertically upward at a rate of 250 ft/s, v₀ = 250 ft/s

The projectile formula is given by :

$h=-16t^2+v_ot$

We need to find the time(s), in seconds, the arrow is at a height of 500 ft. So,

$-16t^2+250t=500$

On solving the above quadratic equation, we get the value of t as, t = 2.35 seconds

So, the arrow is at a height of 500 feet at time t = 2.35 seconds. Hence, this is the required solution.

6 0

3 years ago

Read 2 more answers

(1) Predator-prey is the most important type of ecological relationship.

DIA [1.3K]

1) True 2) True 3)True I hope I helped

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2 years ago

A merry-go-round of radius R, shown in the figure, is rotating at constant angular speed. The friction in its bearings is so sma

mel-nik [20]

The angular speed of the merry-go-round reduced more as the sandbag is

placed further from the axis than increasing the mass of the sandbag.

The rank from largest to smallest angular speed is presented as follows;

[m = 10 kg, r = 0.25·R]

${}$ ⇩

[m = 20 kg, r = 0.25·R]

${}$ ⇩

[m = 10 kg, r = 0.5·R]

${}$ ⇩

[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]

${}$ ⇩

[m = 10 kg, r = 1.0·R]

Reasons:

The given combination in the question as obtained from a similar question online are;

1: m = 20 kg, r = 0.25·R

2: m = 10 kg, r = 1.0·R

3: m = 10 kg, r = 0.25·R

4: m = 15 kg, r = 0.75·R

5: m = 10 kg, r = 0.5·R

6: m = 40 kg, r = 0.25·R

According to the principle of conservation of angular momentum, we have;

$I_i \cdot \omega _i = I_f \cdot \omega _f$

The moment of inertia of the merry-go-round, $I_m$ = 0.5·M·R²

Moment of inertia of the sandbag = m·r²

Therefore;

0.5·M·R²· $\omega _i$ = (0.5·M·R² + m·r²)· $\omega _f$

Given that 0.5·M·R²· $\omega _i$ is constant, as the value of m·r² increases, the value of $\omega _f$ decreases.

The values of m·r² for each combination are;

Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²

Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²

Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²

Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²

Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²

Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²

Therefore, the rank from largest to smallest angular speed is as follows;

Combination 3 > Combination 1 > Combination 5 = Combination 6 >

Combination 2

Which gives;

[m = 10 kg, r = 0.25·R] > [m = 20 kg, r = 0.25·R] > [m = 10 kg, r = 0.5·R] > [m =

10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R] > [m = 10 kg, r = 1.0·R].

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1 year ago

Which statement describes an action-reaction pair?

Ksju [112]

Answer:

c yan alam konayan e good lock

8 0

2 years ago

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