The orbitals can be found by counting down the ____

Edit question A 37-cm-long wire of linear density 18 g/m vibrating at its second mode, excites the third vibrational mode of a t

Pachacha [2.7K]

Answer:

176.9N

Explanation:

The following data were given

wire length,L=37cm=0.37m

linear density=18g/m

tube length,=192cm=1.92m,

speed of sound,v=343m/s

Since it is an open-closed tube, the second harmonic frequency is expressed as

$f_{3}=3(\frac{v}{4l} )\\f_{3}=3(\frac{343}{4*1.92})\\f_{3}=133.98Hz$

The relationship between the tension, linear density and second harmonic frequency is expressed as

$f_{3}=\frac{1}{2l_{w}}\sqrt{\frac{T}{\alpha } } \\T=(f_{3}*2l_{w})^{2}\alpha \\T=(133.984*2*0.37)^{2}*18*10^{-3}\\T=176.9N$

7 0

3 years ago

A low-luminosity star has a small and narrow ________, whereas a high-luminosity star has a large and wide one.

vova2212 [387]

A low-luminosity star has a small and narrow <u>habitable zone</u>, whereas a high-luminosity star has a large and wide one.

<h3>What is luminosity of a star?</h3>

The radiant power emitted by a light-emitting item over time is measured as luminosity, which is an absolute measure of radiated electromagnetic power (light).

The total quantity of electromagnetic energy released per unit of time by a star, galaxy, or other celestial object is referred to as luminosity in astronomy.

Learn more about low-luminosity star:
brainly.com/question/13912549
#SPJ4

4 0

1 year ago

Car A (mass 1100 kg) is stopped at a traffic light when it is rearended by car B(mass 1400 kg). Both cars then slide with locke

viva [34]

Answer:

Part a)

$v_a = 3.94 m/s$

Part b)

$v_b = 3.35 m/s$

Part C)

$v_b = 6.44 m/s$

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

Explanation:

Part a)

As we know that car A moves by distance 6.1 m after collision under the frictional force

so the deceleration due to friction is given as

$a = -\frac{F_f}{m}$

$a = -\frac{\mu mg}{m}$

$a = - \mu g$

now we will have

$v_f^2 - v_i^2 = 2ad$

$0 - v_i^2 = 2(-\mu g)(6.1)$

$v_a = \sqrt{(2(0.13)(9.81)(6.1)}$

$v_a = 3.94 m/s$

Part b)

Similarly for car B the distance of stop is given as 4.4 m

so we will have

$v_b = \sqrt{2(0.13)(9.81)(4.4)}$

$v_b = 3.35 m/s$

Part C)

By momentum conservation we will have

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1400 v_b = 1100(3.94) + 1400(3.35)$

$v_b = 6.44 m/s$

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

3 0

3 years ago

Explain how soilds non metals are different from solid metals

marysya [2.9K]

Solid metal is all different types of metals or some thing that a magnet can pick up that's a full hard solid

but a non metal is everything not metal that's a solid like plastic a hot glue gun can burn throught plastic and not metal

so in conclusion metal is stronger and thicker than non modal things

7 0

3 years ago

Determine the answer to the equation 30 km/h × 17 h =

Jobisdone [24]

30 km/h * 17 h = 30*17 km/h *h
= 510 km

5 0

3 years ago

The orbitals can be found by counting down the _____.