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2 years ago

the resistance of a wire of length 80cm and of uniform area of cross-section 0.025cmsq., is found to be 1.50 ohm. Calculate spec

ific resistance of wire in SI units, Help please

Physics

1 answer:

vivado [14]2 years ago

3 0

$Specific\ resistance\ =resistivity\\
From\ formula\ on \ resistance:\ R= \frac{pL}{A}\ p-resistivity,\ L-length,\\ A-area\ of\ cross\ section\\
p= \frac{R*A}{L}= \frac{1,5Ohm*0,025*10^{-4}m^2 }{80* 10^{-2}m }=0,0003515625 Ohm*m$

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In a movie, Tarzan evades his captors by hiding under water for many minutes while breathing through a long, thin reed. Assume t

gladu [14]

Answer: 0.98m

Explanation:

P = -74 mm Hg = 9605 Pa = 9709N/m^2

= 9605 kg m/s^2/m^2

density of water: rho = 1 g/cc = 1 (10^-3 kg)/(10^-2 m)^-3 = 1000 kg/m^3

Pressure equation: P = rho g h

h = P/(rho g)

h = (9605 kg/m/s^2) / (1000 kg/m^3) / (9.8 m/s^2)

h = 0.98 m

0.98m is the maximum depth he could have been.

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2 years ago

A string, 0.28 m long and vibrating in its third harmonic, excites an open pipe that is 0.82 m long into its second overtone res

S_A_V [24]

Answer: 98.

Explanation: it has been stated in the question that sound wave in the string and in the pipe resonated at a specific frequency, this simply implies that the frequency of sound wave in the string equals frequency of sound wave in pipe.

Fs = Fp.

The length (l) of the string is 0.28m and it is vibrating at it third harmonic.

The length of stationary wave on a string at third harmonic is given below as

l = 3λ/2

Where λ = wavelength of sound wave in pipe (λs)

By substituting l = 0.28m into the equation above, we have that

0. 28 = 3λs/2

3λs = 0.28 * 2

3λs = 0.56, λs= 0.56/ 3

λs = 0.187m

Thus the wavelength of wave in the string is 0.187m.

Sound from the string in the pipe is produced at the second overtone ( which is the third harmonic).

Therefore the length of air in the pipe at second overtone ( third harmonic) is given below as

l = 5λp/ 4, we need to get the wavelength of sound in the pipe.

Thus

λp = 4*l/5

λp = 4 * 0.82 / 5

λp = 0.656m.

The velocity of sound waves produced in the pipe is 345m/s thus the frequency of sound in the pipe is gotten using the formulae below

V = fpλp

V= velocity of sound in pipe, fp = frequency of sound in pipe, λp= wavelength of sound in pipe

345 = f / 0.656

fp = 525.92Hz.

As stated in the question, the frequency of sound in pipe is the same as that in string (fp = fs = f) , thus to get the velocity of sound wave in string we use the same formulae of

v = fλ

Where f = frequency of sound in pipe = frequency of sound in string = 525.92Hz.

λ = wavelength of sound in string = 0.187m

Thus v = 525.92 * 0.187 = 98.34 which is closest to 98.

3 0

2 years ago

Assuming motion is on a straight path, what would result in two positive components of a vector?

Aliun [14]

Answer: option A) a train moving north of east at an angle of 25°


Explanation:

1) You need to choose your axis. In this case North is vertical and positive, South is vertical and negative, East is horizontal and positive, and West is horizontal and negative.

2) The vector with the two positive components is a vector in the first quadrant (North and East). That is what North of East 25° means.

3) Regarding the other options:

B) a bus moving North of East at an angle of 95°: since the angle is greater tnan 90° the vector is in the second quadrant: its horizontal component is negative.


 C) a boat moving South of West at an angle of 40°: this is in the third quadrant: the two components are negative.

 D) a car moving south of west at an angle of 10°: as in the option C), this is in the third quadrant: the two components are negative.

3 0

2 years ago

Read 2 more answers

A computer that is 87% efficient consumes 375 kWh of energy. How much useful energy does it provide?

tekilochka [14]

Answer:

326.25 kWh

Explanation:

Efficiency of a machine is defined as the ratio of useful energy to that of the energy consumed by the machine.

Here, efficiency is given as 87% and the energy consumed by the computer is 375 kWh.

Efficiency, $\eta=\frac{\textrm{Useful energy}}{\textrm{Energy consumed}}$

Plug in the values of $\eta=0.87$ and 375 kWh for energy consumed. Solve for useful energy. This gives,

Efficiency, $\eta=\frac{\textrm{Useful energy}}{\textrm{Energy consumed}}\\ 0.87=\frac{\textrm{Useful energy}}{375}\\ \textrm{Useful energy}=0.87\times 375=326.25 \textrm{ kWh}$

Therefore, the useful energy provided by the computer is 326.25 kWh.

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2 years ago

A 8.0 n force acts on a 0.70-kg object for 0.50 seconds. by how much does the object's momentum change (in kg-m/s)? (never inclu

inysia [295]

For Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:
$F=ma$
Rewriting the acceleration as the increment of velocity $\Delta v$ in a time $\Delta t$ : $a= \frac{\Delta v}{\Delta t}$ , F becomes
$F=m \frac{\Delta v}{\Delta t}$
But given the definition of momentum: $p=mv$ , then $m \Delta v$ represents the momentum change. So we can rewrite F as
$F= \frac{\Delta p}{\Delta t}$
And re-arranging the formula we can calculate the value of the change in momentum:
$\Delta p = F \Delta t=(8.0 N)(0.50 s)=4 kg m/s$

4 0

2 years ago