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aev [14]
3 years ago
11

The units of density for a rectangular piece of wood are?

Physics
1 answer:
kipiarov [429]3 years ago
6 0

Density has units of [ mass / length³ ] or [ mass / volume ].

The unit doesn't change with the substance.

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A parallel-plate air capacitor is made from two plates 0.190 m square, spaced 0.770 cm apart. It is connected to a 120 V battery
Naddik [55]

Answer:

aaksj

Explanation:

a) the capacitance is given of a plate capacitor is given by:

C = \epsilon_0*(A/d)

Where \epsilon_0 is a constant that represents the insulator between the plates (in this case air, \epsilon_0 = 8.84*10^(-12) F/m), A is the plate's area and d is the distance between the plates. So we have:

The plates are squares so their area is given by:

A = L^2 = 0.19^2 = 0.0361 m^2

C = 8.84*10^(-12)*(0.0361/0.0077) = 8.84*10^(-12) * 4.6883 = 41.444*10^(-12) F

b) The charge on the plates is given by the product of the capacitance by the voltage applied to it:

Q = C*V = 41.444*10^(-12)*120 = 4973.361 * 10^(-12) C = 4.973 * 10^(-9) C

c) The electric field on a capacitor is given by:

E = Q/(A*\epsilon_0) = [4.973*10^(-9)]/[0.0361*8.84*10^(-12)]

E = [4.973*10^(-9)]/[0.3191*10^(-12)] = 15.58*10^(3) V/m

d) The energy stored on the capacitor is given by:

W = 0.5*(C*V^2) = 0.5*[41.444*10^(-12) * (120)^2] = 298396.8*10^(-12) = 0.298 * 10 ^6 J

6 0
3 years ago
Read 2 more answers
A fire helicopter carries a 700-kg bucket of water at the end of a 20.0-m long cable. Flying back from a fire at a constant spee
rodikova [14]

Answer:

F = 50636.873 N

Explanation:

given,

bucket of water =  700-kg

length of cable = 20 m

Speed  = 40 m/s

angle of the cable = 38.0°

let air resistance be = F

tension in rope be = T

T cos 38° = m×g..................(1)

T sin 38^0= \dfrac{mv^2}{l} + F..........(2)

equation (1)/(2)

tan 38^0 =\dfrac{\dfrac{mv^2}{l} + F}{mg}

       0.781=\dfrac{\dfrac{700\times 40^2}{20} + F}{700\times 9.8}

           F = 50636.873 N

Hence the force exerted on the bucket is equal to F = 50636.873 N

5 0
3 years ago
The overall length of a piccolo is 32.0 cm. The resonating air column vibrates as in a pipe that is open at both ends. (a) Find
lana66690 [7]

Answer:

lowest frequency = 535.93 Hz

distance  between adjacent anti nodes is 4.25 cm

Explanation:

given data

length L = 32 cm = 0.32 m

to find out

frequency and distance between adjacent anti nodes

solution

we consider here speed of sound through air at room temperature 20 degree is  approximately  v = 343 m/s

so

lowest frequency will be = \frac{v}{2L}   ..............1

put here value in equation 1

lowest frequency will be = \frac{343}{2(0.32)}

lowest frequency = 535.93 Hz

and

we have given highest frequency f = 4000Hz

so

wavelength =  \frac{v}{f}   ..............2

put here value

wavelength =  \frac{343}{4000}  

wavelength = 0.08575 m

so distance =  \frac{wavelength}{2}   ..............3

distance =  \frac{0.08575}{2}  

distance = 0.0425 m

so distance  between adjacent anti nodes is 4.25 cm

3 0
3 years ago
Which of these would be an example of unbalanced forces?
mylen [45]

A snowball picks up speed as it rolls down the mountain.<em> (D)</em>

Since the description includes acceleration ("picks up speed"), we know that the forces on the snowball must be unbalanced.

3 0
3 years ago
Read 2 more answers
Calculate the de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s (about 60 mi/h)
PilotLPTM [1.2K]

The de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s is 4.55×10⁻³⁵ m.

<h3>De Broglie wavelength:</h3>

The wavelength that is incorporated with the moving object and it has the relation with the momentum of that object and mass of that object. It is inversely proportional to the momentum of that moving object.

λ=h/p

Where, λ is the de Broglie wavelength, h is the Plank constant, p is the momentum of the moving object.

Whereas, p=mv, m is the mass of the object and v is the velocity of the moving object.

Therefore, λ=h/(mv)

λ=(6.63×10⁻³⁴)/(0.56×26)

λ=4.55×10⁻³⁵ m.

The de Broglie wavelength associated with the object weight 0.56 kg moving with the velocity of 26 m/s is λ=4.55×10⁻³⁵ m.

Learn more about de Broglie wavelength on

brainly.com/question/15330461

#SPJ1

6 0
1 year ago
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