Answer:
hello your question has some missing parts below is the complete question
and the missing diagram
The two speakers emit sound that is 180° out of phase and of a single frequency,ƒ, Find the lowest two frequencies that produce a maximum sound intensity at the positions of Moe and Curly.
answer : 1316.2 hertz
Explanation:
The frequency that produce the maximum sound intensity can be calculated using the relation below
dsin ∅ = n <em>A</em>
where <em>A = </em>dsin ∅ / n when n = 1 . d = 0.800
<em>A</em> = 0.800 * ( 1 / 3.162 )
<em>A</em> = 0.253 m
speed of sound = 333 m/s
frequency = speed /<em> A</em>
<em>= </em>333 / 0.253 = 1316.2 hertz
Answer:
its 1/2 the mass of the object times by its velocity ^ 2
<span>We'll use the momentum-impulse theorem. The x-component of the total momentum in that direction is given by p_(f) = p_(1) + p_(2) + p_(3) = 0.
So p_(1x) = m1v1 = 0.2 * 2 = 0.4 Also p_(2x) = m2v2 = 0 and p_(3x) = m3v3 = 0.1 *v3 where v3 is unknown speed and m3 is the mass of the third particle with the unknown speed
Similarly, the 235g particle, y-component of the total momentum in that direction is given by p_(fy) = p_(1y) + p_(2y) + p_(3y) = 0.
So p_(1y) = 0, p_(2y) = m2v2 = 0.235 * 1.5 = 0.3525 and p_(3y) = m3v3 = 0.1 * v3 where m3 is third particle mass.
So p_(fx) = p_(1x) + p_(2x) + p_(3x) = 0.4 + 0.1v3; v3 = 0.4/-0.1 = - 4
Also p_(fy) = 0.3525 + 0.1v3; v3 = - 0.3525/0.1 = -3.525
So v_3x = -4 and v_3y = 3.525.
The speed is their resultant = âš (-4)^2 + (-3.525)^2 = 5.335</span>
<u>Answer:</u>
<em>Newtons II law: </em>
<em> </em>It is defined as<em> "the net force acting on the object is a product of mass and acceleration of the body"</em> . Also it defines that the <em>"acceleration of an object is dependent on net force and mass of the body".</em>
Let us assume that,a string is attached to the cart, which passes over a pulley along the track. At another end of the string a weight is attached which hangs over the pulley. The hanging weight provides tension in the spring, and it helps in accelerating the cart. We assume that the string is massless and no friction between pulley and the string.
Whenever the hanging weight moves downwards, the cart will accelerate to right side.
<em>For the hanging weight/mass</em>
When hanging weight of mass is m₁ and accelerate due to gravitational force g.
Therefore we can write F = m₁ .g
and the tension acts in upward direction T (negetive)
Now, Fnet = m₁ .g - T
= m₁.a
So From Newtons II law<em> F = m.a</em>
Answer:
The work done is 360 J.
Explanation:
Given that,
Mass = 50 kg
Distance =3 m
We need to calculate the work done
The work done is equal to the product of force and displacement.
Using formula of work done
Where, F = force
D = distance
θ = Angle between force and displacement
Put the value into the formula
Hence, The work done is 360 J.
