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Rashid [163]
2 years ago
13

A 4.0 g string, 0.36 m long, is under tension. The string produces a 500 Hz tone when it vibrates in the third harmonic. The spe

ed of sound in air is 344 m/s. In this situation, the wavelength of the standing wave in the string, in SI units, is closest to:
Physics
1 answer:
almond37 [142]2 years ago
5 0

Answer:

\lambda = 0.24 m

Explanation:

The string vibrates in the third harmonics, n = 3

Length of the string, l = 0.36 m

Frequency of the tone produced, f = 500 Hz

The speed of sound in air is 344 m/s

Calculate the speed of sound produced by the string in the third harmonics:

The frequency of sound is given by the formula:

f = \frac{nv}{2l} \\500 = \frac{3v}{2*0.36}\\500 * 2 * 0.36 = 3v\\v = 360/3\\v = 120 m/s\\v = \lambda f\\\lambda = v/f\\\lambda = 120/500\\\lambda = 0.24 m

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A pyrotechnical releases a 3 kg firecracker from rest. at t=0.4 s, the firecracker is moving downward with a speed 4 m/s. At the
olga2289 [7]

Answer:

a) F = 30 N, b)   I = 12 N s , c)  I = -12 N s , d) ΔI = 0 N s

Explanation:

This exercise is a case at the moment, let's define the system formed by the firecracker and its two parts, in this case the forces during the explosion are internal and the moment is conserved

Initial, before the explosion

     p₀ = m v

The speed can be found by kinematics

     v = v₀ - g t

     v = 0 - 10 0.4

     v = -4.0 m / s

Final after division

     pf = m₁ v₁f + m₂ v₂f

    p₀ = pf

    M v = m₁ v₁f + m₂ v₂f

Where M is the initial mass (M = 3 kg), m₁ is the mass mtop (m₁ = 1 kg) and m₂ in the mass m botton (m₂ = 2kg) and the piece that moves up (v₁f = 6m/s )

a) before the explosion the only force acting on the body is gravity

     F = mg

     F = 3 10 = 30 N

b) The expression for momentum is

     I = Ft

Before the explosion the only force that acts is the weight

    I = mg t

    I = 3 10 0.4

    I = 12 N s

c) To calculate this part we use the conservation of the moment and calculate the speed of the body that descends body 2

    M v = m₁ v₁f + m₂ v₂f

    v₂f = (M v - m₁ v₁f) / m₂

    v₂f = (3 (-4) - 1 6) / 2

   v₂f = - 9 m / 2

The negative sign indicates that body 2 (botton) is descending

Now we can use the momentum and momentum relationship for the body during the explosion

    I = F t = Dp

   F t = pf –po)

   F t= [m₁ v₁f + m₂ v₂f]

   

   I = [1 6 + 2 (-9) -0]

   I = -12 N s

This is the impulse during the explosion the negative sign indicates that it is headed down

d) impulse change

I₀ = Mv

I₀ = 3 *4

I₀ =-12 N s

 ΔI =If – I₀  

ΔI = - 12 – (-12)

ΔI = -0 N s

3 0
2 years ago
A body is projected upward at an angle of 30 degree to the horizontal at an initial speed of 200ms-.In how many seconds will it
Crazy boy [7]

Answer:

20.41 s

3534.80 m

Explanation:

<em><u>In how many seconds will it reach the ground?</u></em>

We are given the initial velocity of the body, which is 200 m/s at a 30° angle.

We know the acceleration in the vertical direction is -9.8 m/s², assuming that the upwards/right direction is positive and the downwards/left direction is negative.

Since we are using acceleration in the y-direction, let's use the vertical component of the initial velocity.

  • 200 · sin(30) m/s

Let's use the fact that at the top of its trajectory, the body will have a final velocity of 0 m/s.

Now we have one missing variable that we are trying to solve for: time t.

Find the constant acceleration equation that contains v₀, v, a, and t.

  • v = v₀ + at

Substitute known values into the equation.

  • 0 = 200 · sin(30) + (-9.8)t
  • -200 · sin(30) = -9.8t
  • t = 10.20408163

Recall that this is only half of the body's trajectory, so we need to double the time value we found to find the total time the body is in the air.

  • 2t = 20.40816327

The body will reach the ground in 20.41 seconds.

<em><u>How far from the point of projection would it strike? </u></em>

We want to find the displacement in the x-direction for the body.

Let's find the constant acceleration equation that contains time t, that we just found, and displacement (Δx).

  • Δx = v₀t + 1/2at²

Substitute known values into the equation. Remember that we want to use the horizontal component of the initial velocity and that the acceleration in the x-direction is 0 m/s².

  • Δx = (200 · cos(30) · 20.40816327) + 1/2(0)(20.40816327)²
  • Δx = 3534.797567

The body will strike 3534.80 m from the point of projection.

4 0
1 year ago
If Emily throws the ball at an angle of 30∘ below the horizontal with a speed of 14m/s, how far from the base of the dorm should
liubo4ka [24]

Emily throws the ball at 30 degree below the horizontal

so here the speed is 14 m/s and hence we will find its horizontal and vertical components

v_x = 14 cos30 = 12.12 m/s

v_y = 14 sin30 = 7 m/s

vertical distance between them

\delta y = 4 m

now we will use kinematics in order to find the time taken by the ball to reach at Allison

\delat y = v_y *t + \frac{1}{2} at^2

here acceleration is due to gravity

a = 9.8 m/s^2

now we will have

4 = 7 * t + \frac{1}{2}*9.8 * t^2

now solving above quadratic equation we have

t = 0.44 s

now in order to find the horizontal distance where ball will fall is given as

d = v_x * t

here it shows that horizontal motion is uniform motion and it is not accelerated so we can use distance = speed * time

d = 12.12 * 0.44 = 5.33 m

so the distance at which Allison is standing to catch the ball will be 5.33 m

8 0
2 years ago
Two balls with charges +Q and +4Q are separated by 3R. Where should you place another charged ball Q0 on the line between the tw
azamat

Answer:

yes independent of the sign or valve of Q

Explanation:

7 0
1 year ago
Which changes would result in a decrease in the gravitational force between two objects? Check all that apply.
Rufina [12.5K]

Answer: 1. decreasing the mass of both objects

2. decreasing the mass of one of the objects

3. increasing the distance between the objects

Explanation: Hope that helped! (:

4 0
1 year ago
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