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3 years ago

Is the classification for an instrument that produces sound by the vibration of a column of air?

Physics

2 answers:

AleksandrR [38]3 years ago

8 0

You have not provided the answers, however if Wind is a choice that is the correct answer

Answer: Wind

earnstyle [38]3 years ago

6 0

<h3>Aerophone is the crorect answer. </h3>

(Apex)

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monochromatic light of wavelength 500 nm is incident normally on a diffraction grating. if the third order maximum is 32. how ma

Alona [7]

Answer:

The total number of maxima that can be seen is 11

Explanation:

Given the data in the question

wavelength λ = 500 nm = 5 × 10⁻⁷ m

if the third order maximum is 32

i.e m = 3 and θ = 32°

Now, we know that condition for diffraction maximum is as follows;

d × sinθ = m × λ

so we substitute in our given values

d × sin( 32° ) = 3 × 5 × 10⁻⁷ m

d × sin( 32° ) = 1.5 × 10⁻⁶ m

d = [ 1.5 × 10⁻⁶ m ] / sin( 32° )

d = 2.83 × 10⁻⁶ m

Now, maxima n when θ = 90° will be;

sin( 90° ) = nλ / d

1 = nλ / d

d = nλ

n = d / λ

we substitute

n = [ 2.83 × 10⁻⁶ m ] / [ 5 × 10⁻⁷ m ]

n = 5.66

so 5 is the max value

hence, total maxima value is;

⇒ 2n + 1 = 2( 5 ) + 1 = 10 + 1 = 11

Therefore, total number of maxima that can be seen is 11

5 0

3 years ago

A 16 N force is applied to an object and 96 J of work is done. How far was the object moved?

ipn [44]

Your answer would be A. You divide 96 by 16 to find the answer

3 0

4 years ago

Read 2 more answers

A ball of mass m = 0.1kg is connected to a rope of length L = 1.2 m. The ball is swung around in a vertical circle and ball is m

bonufazy [111]

Answer:

The speed of the ball is approximately 5.94 m/s

The Tension of the string at the bottom is 3.92 N

Explanation:

We need to find the speed of the ball, which is constant due to the fact that we are in a uniform circular motion. Notice as well that the speed of the ball is the magnitude of the tangential velocity " $v_t$ " (vector that changes direction with the position of the ball but doesn't change magnitude in this case).

We analyze first the top position of the circular motion, for which information on the tension of the string is given (see first free body diagram in the attached picture). We are told that the tension at the top of the movement equals twice the force of gravity on the ball's mass: T - 2*m*g = 1.96 N. And we know that there are two forces acting on the ball in that position (illustrated with the green arrows pointing down): one is the ball's weight due to gravity, and the other is the string's tension. So we can write Newton's second law for this situation:

$F_{net}= T_{top}+W\\F_{net}=2\,W+W\\F_{net}=3\,W\\F_{net}=2.94 N\\$

Newton's second law tells us that the net force should equal the mass of the ball times its acceleration (F = m * a), and in this motion, the acceleration is the centripetal acceleration. Therefore weuse this equation to solve for the centripetal acceleration of the ball:

$m\,a_c=2.94\,N\\a_c=\frac{2.94\,N}{0.1\,kg} \\a_c=29.4\,\frac{m}{s^2}$

The centripetal acceleration is defined as the square of the tangential velocity divided the radius of the circular motion. Then we use it to derive the magnitude of the tangential velocity (speed of the ball):

$a_c=\frac{v^2}{R} \\29.4\,\frac{m}{s^2} =\frac{v_t^2}{R} \\v_t^2=29.4\,(1.2)\,\frac{m^2}{s^2} \\v_t=5.94\,\frac{m}{s}$

So we have found the speed of the ball.

Now we focus our attention to the bottom of the motion, and again use Newton's second law to solve for the string tension (see second free body diagram in the attached picture).

We notice here that the tension and the weight are acting in opposite directions, so we have such into account when finding the net force on the ball, and then solve for the tension knowing the value of the centripetal acceleration (recall that the magnitude of the tangential velocity is the same because of the uniform circular motion).

$F_{net}= T_{bot}-W\\m\,a_c=T_{bot}-0.98\,N\\2.94\,N=T_{bot}-0.98\,N\\T_{bot}=(2.94+0.98)N\\T_{bot}=3.92\,N$