A steel guitar string with a diameter of 0.300 mm and a length of 70.0 cm is stretched by 0.500 mm while being tuned. How much f

Answer:

d = 2021.6 km

Explanation:

This distance problem can be solved using vector analysis; it's best to find each plane's position components before applying the Pythagorean theorem to calculate the separation between them.

For Airplane 1:

Height   y₁ = 800m

Angle θ = 25°

           cos 25 = x / r

           sin 25 = z / r

           x₁ = r cos 20

           z₁ = r sin 25

          x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

          z₁ = 18 103 sin 25 = 7,607 103 m = 7607 m

For Plane 2:

Height   y₂ = 1100 m

Angle θ = 20°

          x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

          z₂ = 20 103 sin 25 = 8.452 103 m = 8452 m

To determine the distance between the planes using the Pythagorean theorem:

         d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Now, we perform the calculations:

        d² = (18126-16314)²  + (1100-800)² + (8452-7607)²

        d² = 3,283 106 + 9 104 + 7,140 105

        d² = (328.3 + 9 + 71.40) 10⁴

        d = √(408.7 10⁴)

        d = 20,216 10² m

        d = 2021.6 km

Response:

A=0.199

Clarification:

We know that  

Mass of spring=m=450 g=

Where 1 kg=1000 g

Frequency of oscillation=

Energy for oscillation is 0.51 J

To determine the amplitude of oscillations.

Energy for oscillator=

Where =Angular frequency

A=Amplitude

Using the formula

Therefore, the amplitude of oscillation=A=0.199

To address this question, we will utilize concepts linked to centripetal force, aligning it with the static frictional force acting on the object. Using this relationship, we can derive the velocity and input the known values. The defined values are:

The maximum velocity can be determined using centripetal force,

Should be equal to,

As a result, the highest speed achievable through the arc without slipping is 9.93m/s

The magnetic field is calculated to be -6.137 × T. Explanation: Given the radio wave wavelength of λ = 0.3 m and an intensity of I = 45 W/m² at times t = 0 and t = 1.5 ns, we determine Bz at the origin. We use the intensity formula relating to the electric field, which incorporates the known intensity of 45, the speed of light c = 3 × m/s², and ∈o as 8.85 × C²/N.m², leading us to E = 184.15. Consequently, applying the equations, we find B = -6.137 × T at the z-axis.

Response:

Once it has crossed, the locomotive requires 17.6 seconds to achieve a speed of 32 m/s.

Details:

  The locomotive's acceleration is 1.6

  The duration taken to pass the crossing is 2.4 seconds.

  We can apply the motion equation, v = u + at, where v represents final velocity, u indicates initial velocity, a denotes acceleration, and t signifies time.

  When the speed reaches 32 m/s, we have v = 32 m/s, u = 0 m/s, and a= 1.6 .

   32 = 0 + 1.6 * t

    t = 20 seconds.

  Therefore, the locomotive attains a speed of 32 m/s after 20 seconds, and it passes the crossing in 2.4 seconds.

Thus, after clearing the crossing, it takes an additional 17.6 seconds to reach the speed of 32 m/s.