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kotykmax [81]
2 years ago
6

A segment of wire of total length 2.0 m is formed into a circular loop having 5.0 turns. If the wire carries a 1.2-A current, de

termine the magnitude of the magnetic field at the center of the loop.
Physics
1 answer:
docker41 [41]2 years ago
8 0

Answer:

Magnetic field at the center of the loop B=5.89\times 10^{-5}\ T.

Explanation:

It is given that total length of wire is 2 m and number of circular loop is 5 turns.

Therefore ,

5\times ( 2\pi r)=2 \ m .\\\\r=\dfrac{1}{5 \pi}=0.064\ m.

We know , magnetic field at the center of loop is given by :

B=N\dfrac{\mu_o i}{2r}

Putting all values in above equation we get :

B=5\times \dfrac{4\pi\times 10^{-7}\times 1.2}{2\times 0.064}\\\\B=5.89\times 10^{-5}\ T.

Hence , this is the required solution.

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A dog is 60m away while moving at constant velocity of 10m/s towards you. Where is the dog after 4 seconds?
evablogger [386]
20m away

the dog was 60m away from. you subtract 40m since it is 10m/s x 4 seconds
4 0
2 years ago
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A boy and a girl are riding a merry-go-round which is turning at a constant rate. the boy is near the outer edge, while the girl
scZoUnD [109]
I think that the girl has greater tangential acceleration because she is closer to the center and the acceleration is greater there. 
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2 years ago
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A 450g mass on a spring is oscillating at 1.2Hz. The totalenergy of the oscillation is 0.51J. What is the amplitude.
Volgvan

Answer:

A=0.199

Explanation:

We are given that  

Mass of spring=m=450 g==\frac{450}{1000}=0.45 kg

Where 1 kg=1000 g

Frequency of oscillation=\nu=1.2Hz

Total energy of the oscillation=0.51 J

We have to find the amplitude of oscillations.

Energy of oscillator=E=\frac{1}{2}m\omega^2A^2

Where \omega=2\pi\nu=Angular frequency

A=Amplitude

\pi=\frac{22}{7}

Using the formula

0.51=\frac{1}{2}\times 0.45(2\times \frac{22}{7}\times 1.2)^2A^2

A^2=\frac{2\times 0.51}{0.45\times (2\times \frac{22}{7}\times 1.2)^2}=0.0398

A=\sqrt{0.0398}=0.199

Hence, the amplitude of oscillation=A=0.199

4 0
2 years ago
Two small aluminum spheres, each of mass 0.0250 kilograms, areseparated by 80.0 centimeters.
sineoko [7]

Answer:

Total number of electrons

N = 7.25 \times 10^{24}

electrons removed from each sphere

N = 5.27 \times 10^{15}

Fraction of electrons transferred is given as

f = 7.27 \times 10^{-10}

Explanation:

As we know that moles is defined as

n =\frac{mass}{molar mass}

n = \frac{0.0250}{0.026982}

n = 0.93

so number of atoms of Al in each sphere is given as

N = 0.93(6.02 \times 10^{23})

N = 5.58 \times 10^{23}

Now number of electrons in each atom is given as

atomic number = number of electrons in each atom = 13

total number of electrons in each sphere is

N = 13 \times (5.58 \times 10^{23})

N = 7.25 \times 10^{24}

Also we know that force of attraction between them is given as

F= \frac{kq_1q_2}{r^2}

1.00 \times 10^4 = \frac{(9\times 10^9)q^2}{0.80^2}

q = 8.4 \times 10^{-4} C

now we have

q = Ne

8.4 \times 10^{-4} = N(1.6 \times 10^{-19}

N = \frac{(8.4 \times 10^{-4})}{1.6 \times 10^{-19}}

N = 5.27 \times 10^{15}

Fraction of electrons transferred is given as

f = \frac{5.27 \times 10^{15}}{7.25 \times 10^{24}}

f = 7.27 \times 10^{-10}

6 0
2 years ago
A roadway for stunt drivers is designed for racecars moving at a speed of 40 m/s. A curved section of the roadway is a circular
Fynjy0 [20]

Answer:

Bank angle = 35.34o

Explanation:

Since the road is frictionless,

Tan (bank angle) = V^2/r*g

Where V = speed of the racing car in m/s, r = radius of the arc in metres and g = acceleration due to gravity in m/s^2

Tan ( bank angle) = 40^2/(230*9.81)

Tan (bank angle) = 0.7091

Bank angle = tan inverse (0.7091)

Bank angle = 35.34o

3 0
2 years ago
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