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4 years ago

Two bar magnets are hung from metal rods and held in the position shown below.

Physics

2 answers:

Flauer [41]4 years ago

7 0

Diagram C would best predict their final position because the north pole is directly next to a south pole and therefore they will both be attracted to each other, pulling the poles (N and S) closer.

fiasKO [112]4 years ago

5 0

The correct diagram is C.

In fact, the magnetic force works as the electrostatic force: like poles repel each other, while opposite poles attract each other. In this example, we have two magnets one next to each other, and we see that the north pole of one magnet is close to the south pole of the other: therefore, the two poles attract each other, so the two magnets will move closer, as in diagram C.

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As altitude increases, air pressure ____.

MAVERICK [17]

Answer:

decreases

Explanation:

Decreases

5 0

4 years ago

Two wooden boxes of equal mass but different density are held beneath the surface of a large container of water. Box A has a sma

Helga [31]

Answer:

Which box has the greater acceleration?

E. Box A

Explanation:

The question is incomplete:

Which box has the greater acceleration?

The bouyant force exerted by the water is equal in both boxes, because it depends on the volume displaced (that is the same for both boxes) and the density of the water.

But, the weight of each boxes is different, according to their density.

For the Box A the acceleration will be:

$m_aa_a=gV(\rho_w-\rho_a)\\\\\rho_aVa_a=gV(\rho_w-\rho_a)\\\\a_a=g\frac{(\rho_w-\rho_a)}{\rho_a}$

The same applies for the Box B:

$a_b=g\frac{(\rho_w-\rho_b)}{\rho_b}$

If we express the ratio of the accelerations, we have:

$a_a/a_b=\frac{(\rho_w-\rho_a)}{\rho_a}*\frac{\rho_b}{(\rho_w-\rho_b)}\\\\$

$a_a/a_b=\frac{(\rho_w-\rho_a)}{(\rho_w-\rho_b)} \frac{\rho_b}{\rho_a}$

We know that both densities are lower than water, because they accelerate upward to the surface when they are released (if they were more dense than water, they would sink more).

We will treat the densities as relative to water, so it becomes rho_w=1.

If we distribute the product, and know that the density of B is higher than the density of A, and both are higher than the product of the densities, we have:

$\rho_w=1\\\\\frac{a_a}{a_b}=\frac{(1-\rho_a)}{(1-\rho_b)} \frac{\rho_b}{\rho_a}=\frac{\rho_b-\rho_a\rho_b}{\rho_a-\rho_a\rho_b}\\\\\\\rho_b>\rho_a>\rho_a\rho_b>0\\\\\\\frac{a_a}{a_b}=\frac{\rho_b-\rho_a\rho_b}{\rho_a-\rho_a\rho_b}>1\\\\a_a>a_b$

The acceleration of A is higher than the acceleration of B.

7 0

4 years ago

A circular conducting loop with a radius of 0.50 m and a small gap filled with a 10.0-Ω resistor is oriented in the xy-plane. If

saul85 [17]

Answer:

Current, I = 0.153 A

Explanation:

Given that,

Radius of the circular conducting loop, r = 0.5 m

Resistance of the resistor, $R=10\ \Omega$

Magnetic field, B = 1 T

Angle with z axis, $\theta=30^{\circ}$

Magnetic field increases to 10 T in 4 seconds

To find,

Magnitude of current.

Solve,

According to Faraday's law, the induced emf is given by:

$\epsilon=\dfrac{\phi_f-\phi_i}{t}$

$\phi_f\ and\ \phi_i$ are final flux and the initial flux respectively.

$\epsilon=NA\ cos\theta\dfrac{B_f-B_i}{t}$

$\epsilon=1\times \pi (0.5)^2\ cos(30)\dfrac{10-1}{4}$

$\epsilon=1.53\ V$

The magnitude of current can be calculated using the Ohm's law as :

$I=\dfrac{\epsilon}{R}$

$I=\dfrac{1.53}{10}$

I = 0.153 A

Therefore, the magnitude of the current that will be caused to flow in the loop is 0.153 A.

3 0

4 years ago

You’ve had practice calculating the grams of hydrogen gas, but it is also possible to calculate the amount of oxygen gas produce

alekssr [168]

Answer:41.991ml

Explanation:

Equations: 2 H2O → 4H+ + 4e + O2 OXIDATION

2 H+ + 2e → H2 REDUCTION

Electrolysis is the chemical decomposition of compounds when electricity is made to pass through a molten compound or solution.

from the oxidation reaction:

1moles of oxygen requires 4moles of electrons to be discharged at the product

F=96500C/mol

Quantity of charge Q=It

=60*60*0.201A

Q=723.6C

Mole=Q/(F*mole ratio of electron)

Mole= 723.6/(4*96500)

Mole=((1809)/(965000))

M=0.0018746114

M1/M2=V1/V2

1/0.00187=22.4dm^3/V2

V2=22.4*0.00187

V2=0.04199129534dm^3

41.99129534ml

5 0

3 years ago

A coil 4.20 cm radius, containing 500 turns, is placed in a uniform magnetic field that varies with time according to B=( 1.20×1

erma4kov [3.2K]

Explanation:

Given that,

Radius of the coil, r = 4.2 cm

Number of turns in the coil, N = 500

The magnetic field as a function of time is given by :

$B=1.2\times 10^{-2}t+2.6\times 10^{-5}t^4$

Resistance of the coil, R = 640 ohms

We need to find the magnitude of induced emf in the coil as a function of time. It is given by :

$\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=\dfrac{-d(NBA)}{dt}\\\\\epsilon=N\pi r^2\dfrac{-dB}{dt}\\\\\epsilon=N\pi r^2\times \dfrac{-d(1.2\times 10^{-2}t+2.6\times 10^{-5}t^4)}{dt}\\\\\epsilon=N\pi r^2\times (1.2\times 10^{-2}+10.4\times 10^{-5}t^3)\\\\\epsilon=500\pi \times (4.2\times 10^{-2})^2\times (1.2\times 10^{-2}+10.4\times 10^{-5}t^3)\\\\\epsilon=2.77(1.2\times 10^{-2}+10.4\times 10^{-5}t^3)\ V$

Hence, this is the required solution.

4 0

3 years ago

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