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3 years ago

describe how you can use a compass needle to determine the direction of a magnetic field at a given location

1 answer:

8 0

The compass doesn’t give you the value of the net magnetic field, just the direction. So, how do you get the magnitude of a particular field from this? The trick is to assume the value of the Earth’s magnetic field and the direction of the compass. Let’s assume that at this location on the Earth, the magnetic field is pointing directly North with a horizontal component of about 2 x 10-5 T.

Now suppose that I do something to create a magnetic field in a known direction and perpendicular to the horizontal component of the Earth’s magnetic field. Here is an example where I put a current carrying wire right over the compass needle. Since the compass is underneath the wire, the magnetic field due to the wire will be 90° to the Earth’s magnetic field.

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4. Surface waves in an earthquake cause damage and destruction by shaking the ground up and down. They are an example of which t

Kaylis [27]

Standing waves is the perfect answer for this

6 0

2 years ago

What property of an object can you determine using mass and volume?

Eddi Din [679]

The answer is density
because density is equal to the mass/volume

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3 years ago

Help with this question.....

mariarad [96]

Answer:

$g_{moon}=1.67 [m/s^{2} ]$

Explanation:

The weight of some mass is defined as the product of mass by gravitational acceleration. In this way using the following formula we can find the weight.

$w =m*g\\$

where:

w = weight [N]

m = mass = 0.06 [kg]

g = gravity acceleration = 10 [N/kg]

Therefore:

$w=0.06*10\\w=0.6[N]$

By Hooke's law we know that the force in a spring can be calculated by means of the following expression.

$F=W\\F = k*x$

where:

k = spring constant [N/m]

x = deformed distance = 6 [cm] = 0.06 [m]

We can find the spring constant.

$k= F/x\\k=0.6/0.06\\k=10 [N/m]$

Since we use the same spring on the moon and the same mass, the constant of the spring does not change, the same goes for the mass.

$F_{moon}=k*0.01\\F = 10*0.01\\F=0.1[N]$

Since this force is equal to the weight, we can now determine the gravitational acceleration.

$F=m*g_{moon}\\g=F/m\\g = 0.1/0.06\\g_{moon} = 1.67[m/s^{2} ]$

6 0

3 years ago

What is the electric field due to a point charge of 20uC at a distance of 1 meter away from it?

Anettt [7]

The electric field due to a point charge of 20uC at a distance of 1 meter away from it is 180000 $\frac{N}{C}$ .

First, you have to know that the space surrounding a load suffers some kind of disturbance, since a load located in that space will suffer a force. The disturbance that this charge creates around it is called an electric field.

In other words, an electric field exists in a certain region of space if, when introducing a charge called witness charge or test charge, it undergoes the action of an electric force.

The electric field E created by the point charge q at any point P, located at a distance r, is defined as:

$E=K\frac{q}{r^{2} }$