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3 years ago

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Use the figure to determine in what direction the north magnetic pole of the compass will point. What type of magnetic pole is t

he compass pointing toward?

1 answer:

Marat540 [252]3 years ago

8 0

The north magnetic pole of the compass will point north. The compass is pointing toward the south magnetic pole because opposite poles attract while like poles repel. This means that the north magnetic pole of the compass is actually attracted to the south magnetic pole of the Earth as if it was a magnet as well.

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Convert 8.1 kilograms to grams

kirill115 [55]

Answer:

8100 g

Explanation:

8.1 kg × 1000

= 8100 g

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3 years ago

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We want to design a cylindrical vacuum capacitor, with a given radius a for the outer cylindrical shell, that will be able to st

anastassius [24]

Solution :

a). Using Gauss's law :

$E=\frac{Q}{4 \pi \epsilon_0r^2}$ , $$b$ .........(1)

Let $E=E_0,\ r=b$ in equation (1)

Therefore, $Q=4 \pi \epsilon_0b^2E_0$ .............(2)

$V_b-V_a = \int^a_b \vec E. d\vec l$

$=\int^a_b E \ dx$

$=\frac{Q}{4 \pi \epsilon_0} \int^a_b \frac{1}{x^2} \ dx$

$=\frac{Q(a-b)}{4 \pi \epsilon_0 a b}$ ....................(3)

Therefore, $U=\frac{1}{2}Q \Delta V$

$=\frac{1}{2}(4 \pi \epsilon_0 b^2 E_0)\left(\frac{Q(a-b)}{4\pi \epsilon_0 a b}\right)$

$=\frac{4 \pi \epsilon_0}{2a} \ E^2_0 b^3(a-b)$ .............(4)

Now differentiating the equation (4) w.r.t. 'b', we get

$b=\frac{3}{4}a$

Thus the radius for the inner cylinder conductor is $b=\frac{3}{4}a$

b). For the energy storage, substitute the radius in (4), we get

$U = 4 \pi\epsilon_0 \frac{27a^3E^2_0}{512}$

This is the amount of energy stored in the conductor.

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3 years ago

What does a magnetic field look like <br>

Lelechka [254]

Answer:Magnetic fields are invisible, at least usually. But scientists from NASA's Space Sciences Laboratory have made them visible as "animated photographs," using sound-controlled CGI and 3D compositing.

Explanation:

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3 years ago

1. An object on Earth and the same object on the Moon would have a difference in

Feliz [49]

Answers: (1) a. weight, (2)b. Force changes by 2/9, (3)b. movement, (4)a. 40,000 Joules, (5)c. the soil will be 5°C.

<h2>Answer 1: a. weight</h2>

Mass and weight are very different concepts.

Mass is the amount of matter that exists in a body, which only depends on the quantity and type of particles within it. This means mass is an intrinsic property of each body and remains the same regardless of where the body is located.

On the other hand, weight is a measure of the gravitational force acting on an object and is directly proportional to the product of the mass $m$ of the body by the acceleration of gravity $g$ :

$W=m.g$

Then, since the Earth and the Moon have different values of gravity, t<u>he weight of an object in each place will vary</u>, but its mass will not.

<h2>Answer 2: b. Force changes by 2/9</h2>

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

$F=G\frac{m_{1}m_{2}}{r^2}$ (1)

Where:

$F$ is the module of the force exerted between both bodies

$G$ is the universal gravitation constant

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

If we double the mass of one object (for example $2m_{1}$ ) and triple the distance between both (for example $3r$ ). The equation (1) will be rewritten as:

$F=G\frac{2m_{1}m_{2}}{(3r)^2}$ (2)

$F=\frac{2}{9}G\frac{m_{1}m_{2}}{r^2}$ (3)

If we compare (1) and (2) we will be able to see the force changes by 2/9.

<h2>Answer 3: b. movement</h2>

The Work $W$ done by a Force $F$ refers to the release of potential energy from a body that is <u>moved</u> by the application of that force to overcome a resistance along a path.

When the applied force is constant and <u>the direction of the force and the direction of the movement are parallel,</u> the equation to calculate it is:

$W=(F)(d)$

Now, <u>when they are not parallel, both directions form an angle</u>, let's call it $\alpha$ . In that case the expression to calculate the Work is:

$W=Fdcos{\alpha}$

Therefore, pushing on a rock accomplishes no work unless there is movement (independently of the fact that movement is parallel to the applied force or not).

<h2>Answer 4: a. 40,000 Joules</h2>

The Kinetic Energy is given by:

$K=\frac{1}{2}mV^{2}$ (4)

Where $m$ is the mass of the body and $V$ its velocity

For the first case (kinetic energy $K_{1}=10000J$ for a car at $V_{1}=30 mph=13.4112m/s$ ):

$K_{1}=\frac{1}{2}mV_{1}^{2}$ (5)

Finding $m$ :

$m=\frac{2K_{1}}{V_{1}^{2}}$ (6)

$m=\frac{2(10000J)}{(13.4112m/s)^{2}}$ (7)

$m=111.197kg$ (8)

For the second case (unknown kinetic energy $K_{2}$ for a car with the same mass at $V_{2}=60 mph=26.8224m/s$ ):

$K_{2}=\frac{1}{2}mV_{2}^{2}$ (9)

$K_{2}=\frac{1}{2}(111.197kg)(26.8224m/s)^{2}$ (10)

$K_{2}=40000J$ (11)

<h2>Answer 5: c. the soil will be 5°C</h2>

The formula to calculate the amount of calories $Q$ is:

$Q=m. c. \Delta T$ (12)

Where:

$m$ is the mass

$c$ is the specific heat of the element. For water is $c_{w}=1 kcal/g\°C$ and for soil is $c_{s}=0.20 kcal/g\°C$

$\Delta T$ is the variation in temperature (the amount we want to find for both elements)

This means we have to clear $\Delta T$ from (12) :

$\Delta T=\frac{Q}{m.c}$ (13)

For Water:

$\Delta T_{w}=\frac{Q_{w}}{m_{w}.c_{w}}$ (14)

$\Delta T_{w}=\frac{1kcal}{(1kg)(1 kcal/g\°C)}$ (15)

$\Delta T_{w}=1\°C)}$ (16)

For Soil:

$\Delta T_{s}=\frac{Q_{s}}{m_{s.c_{s}}$ (17)

$\Delta T_{s}=\frac{1kcal}{(1kg)(0.20 kcal/g\°C)}$ (18)

$\Delta T_{s}=5\°C)}$ (19)

Hence the correct option is c.

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3 years ago

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What do you call an increase in the magnitude of velocity

Vlad1618 [11]

Any change in the magnitude or direction of velocity is "acceleration".

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3 years ago