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lapo4ka [179]
3 years ago
9

you want to describe the harmonic motion of a swing. you find out that it takes 2 seconds for the swing to complete one cycle. w

hat is the swings period and frequency?
Physics
1 answer:
7nadin3 [17]3 years ago
4 0
If it takes the swing 2 seconds to complete one cycle, then
that's the period of its motion ... 2 seconds.

The frequency is the reciprocal of the period ... 1/2 Hz.
('Hz' means 'per second'.)
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Svetradugi [14.3K]

Answer:

No. 67

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3 years ago
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8 0
3 years ago
3. A coil of 100 turns encloses an area of 100 cm2. It is placed at an angle of 700 with a
sasho [114]

Explanation:

Given that,

Number of turns in the coil, N = 100

Area of the coil, A = 100 cm² = 0.01 m²

It is placed at an angle of 70°.

Magnetic field, B = 0.1 Wb/m²

We need to find the magnetic flux through the coil and the emf is induced in the coil after 10⁻³ s.

Magnetic flux is given by :

\phi =BA\cos\theta\\\\\text{For N turns},\\\phi =NBA\cos\theta \\\\\phi=100\times 0.1\times 0.01\times \cos(70)\\\\=0.034\ Wb

So, the magnetic flux through the coil is 0.1 Wb.

Emf induced in the coil is :

\epsilon=\dfrac{-d\phi}{dt}\\\\=\dfrac{0.034}{10^{-3}}\\\\=34\ V

So, 34V of emf is induced in the coil.

7 0
3 years ago
Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equ
Irina-Kira [14]

Answer:

Magnitude of the net force on q₁-

Fn₁=1403 N

Magnitude of the net force on q₂+

Fn₂= 810 N

Magnitude of the net force on q₃+

Fn₃= 810 N

Explanation:

Look at the attached graphic:

The charges of the same sign exert forces of repulsion and the charges of   opposite sign exert forces of attraction.

Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:

F= (k*q*q)/(d)²

F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N

Magnitude of the net force on q₁-

Fn₁x= 0

Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N

Fn₁=1403 N

Magnitude of the net force on q₃+

Fn₃x= 810- 810 cos 60° = 405 N

Fn₃y= 810*sin 60° = 701.5 N

Fn_{3} = \sqrt{405^{2}+701.5^{2}  }

Fn₃ = 810 N

Magnitude of the net force on q₂+

Fn₂ = Fn₃ = 810 N

6 0
3 years ago
A 4.0 Ω resistor has a current of 3.0 A in it for 5.0 min. How many electrons pass 3. through the resistor during this time inte
musickatia [10]

Answer:

Number of electrons, n=5.62\times 10^{21}

Explanation:

It is given that,

Resistance, R = 4 ohms

Current, I = 3 A

Time, t = 5 min = 300 s

We need to find the number of electrons pass through the resistor during this time interval. Let the number of electron is n.

i.e. q = n e ...............(1)

And current, I=\dfrac{q}{t}

I\times t=n\times e

n=\dfrac{It}{e}

e is the charge of an electron

n=\dfrac{3\ A\times 300\ s}{1.6\times 10^{-19}}

n=5.62\times 10^{21}

So, the number of electrons pass through the resistor is 5.62\times 10^{21}. Hence, this is the required solution.

6 0
3 years ago
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