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irina [24]
3 years ago
7

When the switch in the drawing is closed, the current in the coil increases to its equilibrium value. While the current is incre

asing there is an induced current in the metal ring. The ring is free to move. What happens to the ring? a. It does not move b. It jumps upward c. It jumps downward
Physics
1 answer:
Crazy boy [7]3 years ago
3 0

Answer:

the previous correct answer is b

Explanation:

When the circuit is closed in the system, a current is induced that follows the lenz law, which opposes the change that is occurring and therefore the coil increases and the idicidal current in the ring must reach the maximum oppositing is the current of the coil, so quiet force is repulsion

Consequently, the previous correct answer is b

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The instruction booklet for your pressure cooker indicates that its highest setting is 12.3 psi . you know that standard atmosph
zmey [24]
<span>118 C The Clausius-Clapeyron equation is useful in calculating the boiling point of a liquid at various pressures. It is: Tb = 1/(1/T0 - R ln(P/P0)/Hvap) where Tb = Temperature boiling R = Ideal Gas Constant (8.3144598 J/(K*mol) ) P = Pressure of interest Hvap = Heat of vaporization of the liquid T0, P0 = Temperature and pressure at a known point. The temperatures are absolute temperatures. We know that water boils at 100C at 14.7 psi. Yes, it's ugly to be mixing metric and imperial units like that. But since we're only interested in relative pressure differences, it's safe enough. So P0 = 14.7 P = 14.7 + 12.3 = 27 T0 = 100 + 273.15 = 373.15 And for water, the heat of vaporization per mole is 40660 J/mol Let's substitute the known values and calculate. Tb = 1/(1/T0 - R ln(P/P0)/Hvap) Tb = 1/(1/373.15 K - 8.3144598 J/(K*mol) ln(27/14.7)/40660 J/mol) Tb = 1/(0.002679887 1/K - 8.3144598 1/K ln(1.836734694)/40660) Tb = 1/(0.002679887 1/K - 8.3144598 1/K 0.607989372/40660) Tb = 1/(0.002679887 1/K - 5.055103194 1/K /40660) Tb = 1/(0.002679887 1/K - 0.000124326 1/K) Tb = 1/(0.002555561 1/K) Tb = 391.3034763 K Tb = 391.3034763 K - 273.15 Tb = 118.1534763 C Rounding to 3 significant figures gives 118 C</span>
3 0
3 years ago
A typical electric refrigerator has a power rating of 500 Watts, which is the rate (J/s) at which electrical energy is supplied
Goshia [24]

Answer:

The rate of heat removed from inside the refrigerator is 300 watts.

Explanation:

By the First Law of Thermodynamics and the definition of a Refrigeration Cycle, we have the following formula to determine the rate of heat removed from inside the refrigerator (\dot Q_{L}), in watts:

\dot Q_{L} = \dot Q_{H}-\dot W (1)

Where:

\dot Q_{H} - Rate of heat released to the room, in watts.

\dot W - Rate of electric energy needed by the refrigerator, in watts.

If we know that \dot Q_{H} = 800\,W and \dot W = 500\,W, then the rate of heat removed from inside the refrigerator is:

\dot Q_{L} = \dot Q_{H}-\dot W

\dot Q_{L} = 300\,W

The rate of heat removed from inside the refrigerator is 300 watts.

3 0
2 years ago
What is the relationship between lightning and atoms
Rufina [12.5K]

Answer:

The answer is A

Explanation:

Lightning is formed by electrons in the air

6 0
2 years ago
Which best describes a circuit in series?
Tpy6a [65]
A circuit which only has one path for current to follow
6 0
3 years ago
A=(v – u)/t<br><br> i. V as subject<br> ii. U as subject<br> iii. T as subject
Norma-Jean [14]

Answer:

i.

a =  \frac{(v - u)}{t}  \\ v - u = at \\ v = at + u

ii.

a =  \frac{(v - u)}{t}  \\ v - u = at \\ u = v - at

iii.

a =  \frac{(v - u)}{t}  \\ ta = v - u \\ t  =   \frac{(v - u)}{a}

I hope I helped you^_^

8 0
3 years ago
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