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3 years ago

15

A pair of in-phase stereo speakers is placed side by side, 0.764 m apart. You stand directly in front of one of the speakers, 2.

34 m from the speaker.What is the lowest frequency that will produce constructive interference at your location?
f= ?Hz

1 answer:

Goryan [66]3 years ago

7 0

Answer:

f = 2858.33 Hz

Explanation:

given,

distance between speaker (A) and the person = 2.34 m

Distance between speaker (B) and the person is AB =

= $\sqrt{(0.764)^2 + (2.34)^2}$

= 2.46 m

path difference d = BP - AP

= 2.46 - 2.34 m

= 0.12 m

now, λ = 0.12

speed of sound = 343 m/s

$f= \dfrac{c}{\lambda}$

$f= \dfrac{343}{0.12}$

f = 2858.33 Hz

the lowest frequency that will produce constructive interference is equal to

f = 2858.33 Hz

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a wrench weighs 5.24 newtons on earth. when it is taken to the Moon, where g =1.16 m/s2 how much does it weigh?

Andrew [12]

“Weight of the wrench” on “the moon” is “6.07 kg”.

<u>Explanation</u>:

Weight of the wrench is 5.24 N

Weight of the wrench in kilograms = W × g

Taken “g” on the moon is $1.16 \mathrm{m} / \mathrm{s}^{2}$

$=5.24 \mathrm{N} \times 9.81 \mathrm{m} / \mathrm{s}^{2}=51.352 \mathrm{kg}$

Weight of the wrench in kilograms is 51.352 kg.

Formula to calculate weight of the object on the moon is

$\frac{\text {weight of the object on earth}}{9.81 \mathrm{m} / \mathrm{s}^{2}} \times 1.16 \mathrm{m} / \mathrm{s}^{2}$

Substitute the values given,

$=\frac{51.352 \mathrm{kg}}{9.81 \mathrm{m} / \mathrm{s}^{2}} \times 1.16 \mathrm{m} / \mathrm{s}^{2}$

$=5.234 \times 1.16$

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3 years ago

1. When you shine light with a wavelength of 400 nm at 50% intensity, what is observed?

riadik2000 [5.3K]

Answer:

I think it might be A

Explanation:

the reason I think this answer is because the intensity is 50%.

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3 years ago

Find a) molar fraction, b) concentration in mol / and c) concentration in mol / kg of an aqueous solution with 25% alcohol: the

igor_vitrenko [27]

Explanation:

Answer

( 58.3 mL ) (

0.789 g

1 mL

) (

1 mole

46 g

) = 1 mole C2H5OH

( 500 mL H2O ) (

1 g

1 mL

) (

1 mole

18 g

) = 27.8 mole H2O

The total moles = 1 mole C2H5OH + 27.8 mole H2O = 28.8 moles

The mole fraction =

moles C2H5OH

total moles

=

1 mole C2H5OH

28.8 total moles

= 0.035

The mole percent would be 3.5%.

What is the weight fraction?

Answer

( 58.3 mL ) (

0.789 g

1 mL

) = 46 g C2H5OH

( 500 mL H2O ) (

1 g

1 mL

) = 500 g H2O

The total mass = 46 g C2H5OH + 500 g H2O = 546 g

The mass fraction =

mass C2H5OH

total mass

=

46 g C2H5OH

546 total grams

= 0.084

The mass percent would be 8.4%.

What is the molarity?

Answer

The molarity =

moles C2H5OH

L of solution

=

1 mole C2H5OH

.5583 L

= 1.79 M

What is the molality?

Answer

The molality =

moles C2H5OH

kg of solvent

=

1 mole C2H5OH

0.5 kg H2O

= 2

8 0

3 years ago

A solenoidal coil with 25 turns of wire is wound tightly around another coil with 300 turns. The inner solenoid is 25.0 cm long

joja [24]

Answer:

$5.65487\times 10^{-8}\ Wb$

$1.17\times 10^{-5}\ H$

-0.020475 V

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

$N_1$ = Number of turns of coil = 25

$N_2$ = Number of turns of coil 2 = 300

$\frac{di_2}{dt}$ = Rate of current increased = $1.75\times 10^3\ A/s$

d = Diameter = 2 cm

r = Radius = $\frac{d}{2}=\frac{2}{2}=1\ cm$

A = Area = $\pi r^2$

Magnetic field in the solenoid is given by

$B=\mu_0\frac{N_2}{l}I\\\Rightarrow B=4\pi\times 10^{-7}\frac{300}{0.25}\times 0.12\\\Rightarrow B=0.00018\ T$

Magnetic flux is given by

$\phi=BA\\\Rightarrow \phi=0.00018\times \pi\times 0.01^2\\\Rightarrow \phi=5.65487\times 10^{-8}\ Wb$

The average magnetic flux through each turn of the inner solenoid is $5.65487\times 10^{-8}\ Wb$

Mutual inductance is given by

$L=\frac{N_1\phi}{i_1}\\\Rightarrow L=\frac{25\times 5.65487\times 10^{-8}}{0.12}\\\Rightarrow L=1.17\times 10^{-5}\ H$

The mutual inductance of the two solenoids is $1.17\times 10^{-5}\ H$

Induced emf is given by

$V=-L\frac{di_2}{dt}\\\Rightarrow V=-1.17\times 10^{-5}\times 1.75\times 10^3\\\Rightarrow V=-0.020475\ V$

The emf induced in the outer solenoid by the changing current in the inner solenoid is -0.020475 V

5 0

3 years ago