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3 years ago

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A 55.0-g piece of copper wire is heated, and the temperature of the wire changes from 19.0°C to 86.0°C. The amount of heat absor

bed is 343 cal. What is the specific heat of copper? Show your work.

1 answer:

Alinara [238K]3 years ago

4 0

The specific heat of a metal or any element or compound can be determined using the formula Cp = delta H / delta T / mass. delta pertains to change. That is change in enthalpy and change in temperature. From the given data, Cp is equal to 343 cal per (86-19) c per 55 grams. This is equal to 0.093 cal / g deg. Celsius

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A straight 1.20 m long conductor has a 2.00 A current travelling toward the East. Earth's magnetic field in this location is 4.9

KATRIN_1 [288]

Answer:

Force's magnitude $=1.176\,10^{-4}\,N$

Direction: down (towards the center of the Earth)

Explanation:

Recall that the magnetic force on a conductor of length L carrying a current I in a magnetic field B is given by the equation: $F=I\,L\,B$ in the case the magnetic field B and the direction of the current are at 90 degrees from each other (which is our case). The direction of the force will be given by the "right hand rule" associated with the vector product that defines this force.

Since the current is moving East, and the magnetic field of the Earth goes from North to South, the resultant Force vector will be pointing towards the Earth (and perpendicular to the plane defined by the current's direction and the magnetic field B)

The magnitude of the force, is given by the formula above, and given that all quantities to be considered are is SI units, it will result in Newtons (N):

$F=I\,L\,B\\F=2\,*\,1.2\,*\,4.9\,10^{-5}\,N\\F=1.176\,10^{-4}\,N$

8 0

3 years ago

A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this

wlad13 [49]

Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

$r=r_o\times A^{\frac{1}{3}}$

$r_o=1.25 \times 10^{-15} m$ = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

$r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}$

$r=4.6656\times 10^{-15} m=4.6656 fm$

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) $F=k\times \frac{q_1q_2}{a^2}$

k= $9\times 10^9 N m^2/C^2$ = Coulombs constant

$q_1,q_2$ = charges kept at distance 'a' from each other

F = electrostatic force between charges

$q_1=+1.602\times 10^{-19} C$

$q_2=+1.602\times 10^{-19} C$

Force of repulsion between two protons on opposite sides of the diameter

$a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m$

$F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}$

$F=2.6527 N$

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

6 0

2 years ago

Given the equation E = P/N , solve for P

Mars2501 [29]

Answer:

E=P/N

multiply both sides by N

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2 years ago

The membrane of a living cell can be approximated by a parallel-plate capacitor with plates of area 4.50×10−9 m2 , a plate separ

Marizza181 [45]

The energy stored in the membrane is $6.44\cdot 10^{-14} J$

Explanation:

The capacitance of a parallel-plate capacitor is given by

$C=\frac{k\epsilon_0 A}{d}$

where

k is the dielectric constant of the material

$\epsilon_0$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

For the membrane in this problem, we have

k = 4.6

$A=4.50\cdot 10^{-9} m^2$

$d=8.1\cdot 10^{-9} m$

Substituting, we find its capacitance:

$C=\frac{(4.6)(8.85\cdot 10^{-12})(4.50\cdot 10^{-9})}{8.1\cdot 10^{-9}}=2.26\cdot 10^{-11} F$

Now we can find the energy stored: for a capacitor, it is given by

$U=\frac{1}{2}CV^2$

where

$C=2.26\cdot 10^{-11} F$ is the capacitance

$V=7.55\cdot 10^{-2} V$ is the potential difference

Substituting,

$U=\frac{1}{2}(2.26\cdot 10^{-11} F)(7.55\cdot 10^{-2})^2=6.44\cdot 10^{-14} J$

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3 years ago