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Rashid [163]
2 years ago
13

A 4.0 g string, 0.36 m long, is under tension. The string produces a 500 Hz tone when it vibrates in the third harmonic. The spe

ed of sound in air is 344 m/s. In this situation, the wavelength of the standing wave in the string, in SI units, is closest to:
Physics
1 answer:
almond37 [142]2 years ago
5 0

Answer:

\lambda = 0.24 m

Explanation:

The string vibrates in the third harmonics, n = 3

Length of the string, l = 0.36 m

Frequency of the tone produced, f = 500 Hz

The speed of sound in air is 344 m/s

Calculate the speed of sound produced by the string in the third harmonics:

The frequency of sound is given by the formula:

f = \frac{nv}{2l} \\500 = \frac{3v}{2*0.36}\\500 * 2 * 0.36 = 3v\\v = 360/3\\v = 120 m/s\\v = \lambda f\\\lambda = v/f\\\lambda = 120/500\\\lambda = 0.24 m

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Two 7.0cm×7.0cm metal electrodes are spaced 1.0 mm apart and connected by wires to the terminals of a 9.0 v battery.
AleksandrR [38]

Answer:

q=390.285x10^{-12}C

Explanation:

Those kind of problems of electric physics is about capacitors, so the normal questions are:

What are the charge on each electrode?

Solve this you can get other information required in the problem or write down the other questions you need

7.0cm*\frac{1m}{100cm}=70x10^{-3}m

A=70x10^{-3}m*70x10^{-3}m=4.9x10^{-3}m62

Capacitance

C=E_o*\frac{A}{d}

E_o=8.85x10^{-12}\frac{C^2}{N*m^2}

C=8.85x10^{-12}\frac{C^2}{N*m^2}*\frac{4.9x10^{-3}m^2}{1x10^{-3}m}

C=43.365x10^{-12}F

C=43.365pF

The charge is find by the equation

q=C*V

q=43.365pF*9V

q=390.285x10^{-12}C

5 0
2 years ago
A box is at rest on a ramp at an incline of 22°. The normal force on the box is 538 N.
fomenos

Answer: 580 N

Refer to attached figure.

The angle of inclination is 22 degrees

weight (gravitational force) acts downwards.

Normal force is a contact force which acts perpendicular to the point of contact.

The horizontal component (mg cos 22 ) balances the normal force and the vertical component balances the frictional force.

Gravitational force on an object = mg

The normal force N= mg cos 22

\Rightarrow mg =\frac{N}{cos22}=\frac{538 N}{0.927}=580 N





8 0
2 years ago
Read 2 more answers
Diesel engines give more miles per gallon than gasoline engines, but some of this is due to the higher energy content of diesel
Lina20 [59]

Answer:

0.106

Explanation:

For 1 liter of diesel the car can get 19 km, if it takes 0.2 MJ for each km then it would take the total energy of 19*0.2 = 3.8 MJ to move an aerodynamic car 19 km. Since 1 liter of of diesel also contains 36 MJ in internal energy, then the efficiency of the diesel engine is the ratio of its output energy over its input energy:

\frac{3.8}{36} = 0.106

5 0
2 years ago
Shareen performs a skit to model a method of charging. In the skit, a painter shakes her hand and gets paint on her.
Veseljchak [2.6K]

Answer:

It models conduction because the painter represents a charged object and the paint represents electrons that are transferred through contact.

Explanation:

Conduction phenomenon of charging is the process of charging in which two bodies are made in contact with each other so that charges are transferred due to potential difference of two bodies.

here we know that when hands are shake then it will have paint on it. so here due to hand shake the hands are in contact with charge particles and due to contact the electrons are transferred to the hand.

Now here we need to assume that charge of paint must be opposite that of the charge on the hand because only due to opposite charge attraction the paint must be transferred to the hand

SO here correct answer will be

It models conduction because the painter represents a charged object and the paint represents electrons that are transferred through contact.

5 0
2 years ago
Read 2 more answers
The density of nuclear matter is about 1018 kg/m3. Given that 1 mL is equal in volume to 1 cm3, what is the density of nuclear m
Sonbull [250]

Answer:

density is 10^{6} Mg/µL

Explanation:

given data

density of nuclear = 10^{18} kg/m³

1 ml = 1 cm³

to find out

density of nuclear matter in Mg/µL

solution

we know here

1 Mg = 1000 kg

so

1 m³ is equal to 10^{6} cm³

and here 1 cm³ is equal to  1 mL

so we can say 1 mL is equal to 10³ µL

so by these we can convert density

density = 10^{18} kg/m³

density = 10^{18} kg/m³ × \frac{10^{-3} }{10^{6} }  Mg/µL

density =  10^{6} Mg/µL

8 0
2 years ago
Read 2 more answers
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