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3 years ago

14

The frequency of the longitudinal standing wave shown in the drawing is 440 hz. the tube is open at both ends. what is the funda

mental frequency of the tube?

1 answer:

bogdanovich [222]3 years ago

6 0

440hz is the tube opened at both ends

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26. The ice on the front windshield of the car had formed when moisture condensed during the night. The ice melted quickly after

fredd [130]

Answer:

The answer is "Option B".

Explanation:

Please find the complete question in the attachment file.

When precipitation accumulated and during the night, the ice on the front windshield of the car also formed. Because after the car was heated up the following morning, the ice has melted easily because the thawing vent, that blasts on the front windshield, were switched on full force. Although there was no attempt to refrigerate the passenger door, the ice there dissolved at the very same price as the ice on the front window, quite significantly endangering the legitimacy of the clarification of a speed at which the snow melts.

3 0

3 years ago

A box weighing 2.4 x 10^2 Newton's is lifted at a constant speed to a shelf 1.2 meters high in 4.0 seconds. What power is requir

Anuta_ua [19.1K]

Answer:

72Watts

Explanation:

Power = Force * distance/Time

Given the following

Force = 2.4 x 10^2 = 240N

distance = 1.2m

Time = 4.0secs

Substitute

Power = 240*1.2/4

Power = 60 * 1.2

Power = 72watts

Hence the power required to lift the box is 72Watts

3 0

2 years ago

The furnace keeps houseAat 25◦C, while thefurnace in houseBkeeps it at 20◦C. Which house requires heat to be supplied by its fur

EleoNora [17]

Answer:

House A requires heat at a slightest faster rate than B

Explanation:

House A requires heat at a slightest faster rate than B due to the slight high temperature the furnace A is.

5 0

3 years ago

In 1984 a team of german physicists synthesized a new element by smashing an iron-58 nucleus into a lead-208 target. if a neutro

dimulka [17.4K]

Here in all nuclear reactions we can say that mass conservation and charge conservation is always true

Here iron nuclei smashed into lead nuclei and then a new nuclei will form which will released along with a neutron

Now in this reaction mass and charge will remain conserved

mass number of iron + mass number of lead = mass number of new nuclei + mass number of neutron

58 + 208 = x + 1

x = 265

so the new nuclei formed will have mass number A = 265

now we will use charge conservation

Number of protons in iron + number of protons in lead = number of protons in new nuclei + number of protons in neutron

26 + 82 = z + 0

z = 108

so the new nuclei will form with atomic number z = 108 and mass number A = 265

If we refer periodic table to find such atom we will see that this is $^{265}Hs$

so the new nuclei formed is $^{265}Hs$

5 0

2 years ago

My Notes An electron is released from rest on the axis of a uniform positively charged ring, 0.500 m from the ring's center. If

dsp73

Answer:

Velocity of the electron = v = 1.2\times 10^8\ m/s.

Explanation:

Given,

Mass of the electron = $m_e\ =\ 9\times 10^{-31}\ kg$
Charge on the electron = $q_e\ =\ 1.62\times 10^{-19}\ C$
Charge density of the ring = $\rho\ =\ +1.00\times 10^{-6}\ C/m$
Radius of the ring = R = 0.70 m
Distance between the electron ant the center or the ring = x = 0.5 m

Now total charge on the ring = $Q\ =\ \rho\times 2\pi R$

Potential energy due to the charged ring to the point on the x-axis is

$P.E.\ =\ \dfrac{KQq_e}{\sqrt{R^2\ +\ x^2}}\\$

Let v be the velocity of the electron at the center of the ring.

Total kinetic energy of the electron = $\dfrac{1}{2}m_ev^2\\$

Now, From the conservation of energy,

the total potential energy of the electron at initially is converted to the total kinetic energy of the electron at the center of the ring,

$\therefore P.E.\ =\ K.E.\\\Rightarrow \dfrac{KQ}{\sqrt{R^2\ +\ x^2}}\ =\ \dfrac{1}{2}m_ev^2\\\Rightarrow v\ =\ sqrt{\dfrac{2 KQq_e}{m_e\sqrt{R^2\ +\ x^2}}}\\\Rightarrow v\ =\ \sqrt{\dfrac{2Kq_e\rho \times 2\pi R}{m_e\sqrt{R^2\ +\ x^2}}}\\\Rightarrow v\ =\ \sqrt{\dfrac{2\times 9\times 10^9\times 1.0\times 10^{-6}\times 2\times 3.14\times 0.7\times 1.6\times 10^{-19}}{9\times 10^{-31}\times \sqrt{0.7^2\ +\ 0.5^2}}}\\\Rightarrow v\ =\ 1.2\times 10^8\ m/s.$

Hence the velocity of the electron on the center of the charged ring is $1.2\times 10^8\ m/s.$

5 0

3 years ago