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3 years ago

10

The specific heat of copper is 0.39 j/g*degrees celcius how much heat is needed to raise the temperature of 1000.0 g of copper f

rom 25.0 degrees celsius to 45.0 degrees celcius

1 answer:

lapo4ka [179]3 years ago

6 0

In order to calculate the amount of energy needed, we will apply:
heat energy = mass x specific heat capacity x change in temperature
Q = mcΔT

Substituting the given values,
Q = 1000 x 0.39 x (45 - 25)
Q = 7,800 Joules of heat energy are needed

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The conductive tissues of the upper leg can be modeled as a 40-cm-long, 12-cm-diameter cylinder of muscle and fat. The resistivi

ahrayia [7]

To calculate and solve the problem it is necessary to apply the concepts related to resistance and resistivity.

The equation that is responsible for relating the two variables is:

$R = \rho \frac{L}{A}$

Where,

R= Resistance of the conductor

$\rho =$ Resistivity of the conductor material

L = Length

A = Cross-sectional area of conductor

With the previous values the area of the muscle (Real Muscle-82%)is,

$A_m = (0.82)\pi r^2 = (0.82)\pi (12/2*10^{-2})^2$

$A_m = 9.274*10^{-3}m^2$

Using the equation from Resistance we have that at the muscle the value is:

$R_m = \rho \frac{L}{A}$

$R_m = \frac{13*(0.4)}{9.274*10^{-3}}$

$R_m = 560.70\Omega$

At the same time we can make the same process to calculate the resistance of the fat, then

$A_m = (0.18)\pi r^2 = (0.18)\pi (12/2*10^{-2})^2$

$A_m = 2.0357*10^{-3}m^2$

The resistance of the fat would be,

$R_f = \rho \frac{L}{A}$

$R_f = \frac{25*(0.4)}{2.0357*10^{-3}}$

$R_f = 4912.3\Omega$

Then the total resistance in a set as the previously writen, i.e, in parallel is:

$R=\frac{R_mR_f}{R_m+R_f}$

$R = \frac{(560.70)(4912.3)}{4912.3+560.70}$

$R = 502.62\Omega$

We can here apply Ohm's law, then

$I= \frac{V}{R}$

$I = \frac{1.5}{502.62}$

$I = 2.984*10^{-3}A$

$I = 2.984mA$

3 0

4 years ago

A sound wave with a waveenght of 2.5 meters traves 660 meters in 2 seconds calculate the frequemcy of the wave to cacuation the

finlep [7]

The frequency of the wave is 132 Hz

Explanation:

To calculate the speed of the wave, we can use the following formula:

$v = \frac{d}{t}$

where

d is the distance travelled by the wave

t is the time elapsed

For the sound wave in this problem, we have:

d = 660 m is the distance travelled

t = 2 s is the time interval considered

Substituting and solving for v, we find the speed of the sound wave:

$v=\frac{660}{2}=330 m/s$

Now we can calculate the frequency of the wave by using the wave equation:

$v=f\lambda$

where

v = 330 m/s is the speed of the wave

$\lambda=2.5 m$ is the wavelength

f is the frequency

Solving for f, we find:

$f=\frac{v}{\lambda}=\frac{330}{2.5}=132 Hz$

Learn more about wavelength and frequency:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

7 0

3 years ago

A wave with an amplitude of 9.3 mm is traveling along a string whose linear mass density is 230 g/m and whose tension is 65 N. I

dsp73

Answer:

The rate of transfer of energy is equal to 23.76W or 23.76J/s as may be required both forms are correct. The physical quantities needed to calculate the rate of energy transfer are the linear mass density or mass per unit length, tension force, amplitude, angular frequency( which is equal to 2pi •f )

Explanation:

The required quantity is the average power or average rate of energy transfer which differs from the maximum or instantaneous rate of energy transfer. The calculation steps to the answer above can be found in the attachment below. Should the requested quantity be the instantaneous quantity the answer will be 2 x Pav which equals 47.52W or 47.52J/s.

3 0

3 years ago

Which one of the 4 answers is it?<br><br><br>

slavikrds [6]

5 n to the right I think Hope this helps

8 0

3 years ago

Thomson's experiments accounted for: the mass of the electron the fundamental charge the force on a current-carrying wire the el

gogolik [260]

Thomson experiment he calculated the charge to mass ratio just be passing the fundamental charge through a tube

He calculated the charge to mass ratio just by finding the deflection of charge while it is passing through the constant electric field

so here we will use the deflection as following

let say it passes the field of length "L"

so here we have

$t = \frac{L}{v}$

now in the same time if it deflect by some distance

$\delta y = \frac{1}{2}at^2$

$\delta y = \frac{1}{2}{eE}{m}t^2$

now by solving this equation we can find e/m ratio

so here correct answer will be

the electron's charge-to-mass ratio

7 0

3 years ago

Read 2 more answers