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zaharov [31]
2 years ago
13

A circular coil of five turns and a diameter of 30.0 cm is oriented in a vertical plane with its axis perpendicular to the horiz

ontal component of the Earth's magnetic field. A horizontal compass placed at the coil's center is made to deflect 45.0° from magnetic north by a current of 0.600 A in the coil.(a) What is the horizontal component of the Earth's magnetic field?
Physics
1 answer:
____ [38]2 years ago
5 0

The horizontal component of the magnetic field is 12.6 μT.

The magnetic influence on moving electric currents, electric charges, and magnetic materials is described by a magnetic field, which is a vector field. When a charge moves through a magnetic field, a force that is perpendicular to both its own velocity and the magnetic field operates on it.

The horizontal component of the Earth's magnetic field is perpendicular to the axis of a circular coil with five turns and a diameter of D = 30.0 cm that is vertically orientated.

A coil current of I = 0.600 A causes a horizontal compass to deflect 45.0° from magnetic north when it is positioned in the coil's center.

Let B be the magnetic field and R be the radius of the circular coil.

Then the horizontal component of the Earth's magnetic field is given as:

B(h) = B(coil) = μ₀ NI / 2R

B(h) = (4π × 10⁻⁷ ) (5)(0.6) / 0.3

B(h) = 12.6 μT

Learn more about magnetic field here:

brainly.com/question/14411049

#SPJ4

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Answer:

Time, t = 80 seconds

Explanation:

Given that,

The frequency of the oscillating mass, f = 1.25 Hz

Number of oscillations, n = 100

We need to find the time in which it makes 100 oscillations. We know that the frequency of an object is number of oscillations per unit time. It is given by :

f=\dfrac{n}{t}

t=\dfrac{n}{f}

t=\dfrac{100}{1.25\ Hz}

t = 80 seconds

So, it will make 100 oscillations in 80 seconds. Hence, this is the required solution.

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3 years ago
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2 years ago
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Isotope what is the meaning
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Isotope means that a chemical element that has the same number of protons but neutron number differs.

<u>Explanation:</u>

In isotope, the chemical element differs in neutron and nucleon number. Thus, different isotopes of a single component hold the same place in the periodic table.

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3 0
3 years ago
Kiera, a 330 N girl, steps in water that someone spilt on the floor. The coefficient of kinetic friction between Kiera and the f
shutvik [7]

Answer:

<em>The force of kinetic friction between Kiera and the floor is 9.24 N</em>

Explanation:

<u>Friction Force</u>

When an object is moving and encounters friction in rough surfaces, it loses acceleration and/or velocity because the friction force opposes motion.

The friction force when an object is moving on a horizontal surface is calculated by:

Fr=\mu N

Where μ is the coefficient of static or kinetics friction and N is the normal force.

If no forces other then the weight and the normal are acting upon the y-direction, then the weight and the normal are equal in magnitude:

N = W

Thus, the friction force is:

Fr=\mu W

Kiera, the W=330 N girl steps in water that has a coefficient of friction of μ=0.028 with the floor.

The kinetic friction force is:

Fr = 0.028*330

Fr = 9.24 N

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3 years ago
A satellite is in a circular orbit about the earth (ME = 5.98 x 10^24 kg). The period of the satellite is 1.26 x 10^4 s. What is
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Answer: V=5839.051m/s  

Explanation:

According to the <u>Third Kepler’s Law</u> of Planetary motion:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;:

T=1.26(10)^{4}s is the period of the satellite

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=5.98(10)^{24}kg is the mass of the Earth

a  is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).

On the other hand, the orbital velocity V is given by:

V=\sqrt{\frac{GM}{a}}   (2)

Now, from (1) we can find a, in order to substitute this value in (2):

a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}   (3)

a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}   (4)

a=11705845.57m   (5)

Substituting (5) in (2):

V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{11705845.57m}}   (6)

V=5839.051m/s   (7)  This is the speed at which the satellite travels

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