All categories

3 years ago

What part of a standing sound wave does a musician seek in playing a musical note of a specific pitch?

2 answers:

6 0

The correct answer to the question above is that the magician is seeking the wavelength of the standing wave. The part of a standing sound wave, which is its wavelength, the magician is seeking when playing a musical note of a specific pitch.

Anna35 [415]3 years ago

6 0

That is false. I just took the test and it is not the wavelength. The correct answer is antinode NOT wavelength.

You might be interested in

abruzzese [7]

Answer:

I think they use it at night because they want to avoid involving themselves in road accident while crossing the street. When there are reflective strips on their clothes, drivers do notice them even at far ends of the street since the strips will reflect the rays from the car lights to the driver,hence, the driver notifies that there is a pedestrian crossing and therefore slows down.

3 0

3 years ago

Water (2510 g ) is heated until it just begins to boil. if the water absorbs 5.01×105 j of heat in the process, what was the ini

natka813 [3]

E=energy=5.09x10^5J = 509KJ
M=mass=2250g=2.25Kg 
C=specific heat capacity of water= 4.18KJ/Kg 
ΔT= change in temp= ? 
E=mcΔT 
509=(2.25)x(4.18)xΔT 
509=9.405ΔT 
ΔT=509/9.405=54.1degrees 
Initial temp = 100-54 = 46 degrees 
Hope this helps :)

5 0

3 years ago

A satellite is in a circular orbit 21000 km above the Earth’s surface; i.e., it moves on a circular path under the influence of

mina [271]

Answer:

(orbital speed of the satellite) V₀ = 3.818 km

Time (t) = 4.5 × 10⁴s

Explanation:

Given that:

The radius of the Earth is 6.37 × 10⁶ m; &

the acceleration of gravity at the satellite’s altitude is 0.532655 m/s

We can calculate the orbital speed of the satellite by using the formula:

Orbital Speed (V₀) = √(r × g)

radius of the orbit (r) = 21000 km + 6.37 × 10⁶ m

= (2.1 × 10⁷ + 6.37 × 10⁶) m

= 27370000

= 2.737 × 10⁷m

Orbital Speed (V₀) = √(r × g)

Orbital Speed (V₀) = √(2.737 × 10⁷ × 0.532655 )

= 3818.215

= 3.818 × 10³

= 3.818 Km

To find the time it takes to complete one orbit around the Earth; we use the formula:

Time (t) = 2 π × $\frac{r}{V_o}$

= 2 × 3.14 × $\frac{2.737*10^7}{3.818*10^3}$

= 45019.28

= 4.5 × 10 ⁴ s

6 0

3 years ago

A 1900 kg car rounds a curve of 55 m banked at an angle of 11 degrees? . If the car is traveling at 98 km/h, How much friction f

Nat2105 [25]

Answer:

22000 N

Explanation:

Convert velocity to SI units:

98 km/h × (1000 m / km) × (1 h / 3600 s) = 27.2 m/s

Draw a free body diagram. There are three forces acting on the car. Normal force perpendicular to the bank, gravity downwards, and friction parallel to the bank.

I'm going to assume the friction force is pointed down the bank. If I get a negative answer, that'll just mean it's actually pointed up the bank.

Sum of the forces in the radial direction (+x):

∑F = ma

N sin θ + F cos θ = m v² / r

Sum of the forces in the y direction:

∑F = ma

N cos θ - F sin θ - W = 0

To solve the system of equations for F, first solve for N and substitute.

N = (W + F sin θ) / cos θ

Substituting:

((W + F sin θ) / cos θ) sin θ + F cos θ = m v² / r

(W + F sin θ) tan θ + F cos θ = m v² / r

W tan θ + F sin θ tan θ + F cos θ = m v² / r

W tan θ + F (sin θ tan θ + cos θ) = m v² / r

W tan θ + F sec θ = m v² / r

F sec θ = m v² / r - W tan θ

F = m v² cos θ / r - W sin θ

F = m (v² cos θ / r - g sin θ)

Given that m = 1900 kg, θ = 11°, v = 27.2 m/s, and r = 55 m:

F = 1900 ((27.2)² cos 11 / 55 - 9.8 sin 11)

F = 21577 N

Rounding to two sig-figs, you need at least 22000 N of friction force.

4 0

3 years ago

Write down applications of mechanics

harkovskaia [24]

Answer:

Explanation:

applied Mechanics and its Growing Utilisation of Theoretical Mechanics.\

Structural Engineering.

Hydraulics.

Mechanical Engineering.

External Fluid Dynamics.

Planetary Sciences.

Life Sciences.

7 0

3 years ago