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Ksenya-84 [330]
2 years ago
10

Taylor places a nail on a bar magnet. The nail sticks to the magnet when lifted up off the table. She touches a paperclip to the

nail and it sticks to the nail. Explain what happened to the magnetic domains of the nail before and after touching it to the bar magnet.
Physics
2 answers:
diamong [38]2 years ago
8 0

Taylor places a nail on a bar magnet. The nail sticks to the magnet when lifted up off the table. She touches a paperclip to the nail and it sticks to the nail. Explain what happened to the magnetic domains of the nail before and after touching it to the bar magnet.

Ratling [72]2 years ago
5 0
Prior to touching the bar magnet, the magnetic domains in the nail were pointing in random directions. When Taylor touched the nail to the bar magnet the magnetic fields of the magnetic domains aligned and it became a temporary magnet.
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A closed, rigid container holding 0.2 moles of a monatomic ideal gas is placed over a Bunsen burner and heated slowly, starting
Georgia [21]

Answer:

a) 2250 J

b) 0 J

c) 2250 J

Explanation:

a) Since, the process is isochoric

the change in internal energy

\Delta U = n C_v(T_f-T_i)

Here, n = 0.2 moles

Cv = 12.5 J/mole.K

We have to find T_f so we can use gas equation as

\frac{P_1V_1}{P_2V_2} =\frac{T_i}{T_f}\\Since, V_1=V_2    [isochoric/process]\\\Rightarrow \frac{P_{atm}}{4P_{atm}} = \frac{300}{T_f} \\\Rightarrow T_f = 1200 K

So,  \Delta U= 0.2\times12.5(1200-300)\\=2250 J

b) Since, the process is isochoric no work shall be done.

c) By first law of thermodynamics we have

\Delta U = Q-W\\Since, W = 0\\\Delta U = Q\\Therefore, Q = 2250 J

Since, Q is positive 2250 J of heat will flow into the system.

6 0
2 years ago
A hot–air balloon is moving at a speed of 10 meters/second in the +x–direction. The balloonist throws a brass ball in the +x–dir
IrinaVladis [17]
The ball has an initial speed of 10m/s. This is because it is moving with the balloon. Now the balloonist throws the ball 4m/s with respect to himself, so it means that he gives the ball a extra push of 4m/s, so the total speed is 14m/s. Since it takes 30 seconds to reach the ground, the distance travelled is 14*30=420m.
7 0
2 years ago
The wavelength of green light is 550 nm.
SIZIF [17.4K]

Answer:

(a) momentum of photon is 1.205 x 10⁻²⁷ kgm/s

    velocity of electron is 1323.88 m/s

   momentum of the electron is 1.205 x 10⁻²⁷ kgm/s

(b) momentum of photon is 1.506 x 10⁻²⁷ kgm/s

  velocity of electron is 1654.85 m/s

  momentum of the electron is 1.506 x 10⁻²⁷ kgm/s

(c) The momentum of the photon is equal to the momentum of the electron

Explanation:

(a)

wavelength of green light, λ = 550 nm

momentum of photon is given by;

p = \frac{h}{\lambda}\\\\ p = \frac{6.626 *10^{-34}}{550*10^{-9}}\\\\p = 1.205 *10^{-27} \ kg.m/s

velocity of electron is given by;

P = \frac{h}{\lambda} \\\\mv = \frac{h}{\lambda}\\\\v = \frac{h}{m \lambda}\\\\v =   \frac{6.626 *10^{-34}}{(9.1*10^{-31} )(550*10^{-9})}\\\\v = 1323.88 \ m/s

momentum of the electron is given by;

p = mv

p = (9.1 x 10⁻³¹) (1323.88)

p = 1.205 x 10⁻²⁷ kgm/s

(b)

wavelength of red light, λ = 440 nm

momentum of photon is given by;

p = \frac{h}{\lambda}\\\\ p = \frac{6.626 *10^{-34}}{440*10^{-9}}\\\\p = 1.506 *10^{-27} \ kg.m/s

velocity of electron is given by;

v =   \frac{6.626 *10^{-34}}{(9.1*10^{-31} )(440*10^{-9})}\\\\v = 1654.85 \ m/s

momentum of the electron is given by;

p = mv

p =  (9.1 x 10⁻³¹) (1654.85)

p = 1.506 x 10⁻²⁷ kgm/s

(c) The momentum of the photon is equal to the momentum of the electron.

7 0
2 years ago
A kite is 100m above the ground. If there are 200m of string out, what is the angle between the string and the horizontal? (Assu
belka [17]

Answer:

the answer is 30°

Explanation:

due to:

sin law of sines

\frac{sin 90}{200} =\frac{sin\beta }{100}\\arcsin(100\frac{sin90}{200} )= 30°

3 0
2 years ago
You start with spring that's already been stretched an unknown amount from equilibrium. After stretching it an additional 2.0 cm
maxonik [38]

Answer: 35*10^3 N/m

Explanation: In order to explain this problem we know that the potential energy for spring is given by:

Up=1/2*k*x^2 where k is the spring constant and x is the streching or compresion position from the equilibrium point for the spring.

We  also know that with additional streching of 2 cm of teh spring,  the potential energy is 18J. Then it applied another additional streching of 2 cm and the energy is 25J.

Then the difference of energy for both cases is 7 J so:

ΔUp= 1/2*k* (0.02)^2 then

k=2*7/(0.02)^2=35000 N/m

7 0
2 years ago
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