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2 years ago

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Which particles in an atom are acted upon by the strong force? A. Electrons and neutrons B. Neutrons and protons O C. Protons an

d electrons D. Protons, neutrons, and electrons SUE w

1 answer:

BARSIC [14]2 years ago

6 0

$\large \sf \pmb{B) \: Neutrons \: and \: Protons}$

<em>Neutrons And Protons</em> is a particles in an atom and are acted upon by the strong force.

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A solenoid coil with 22 turns of wire is wound tightly around another coil with 330 turns. The inner solenoid is 21.0 cm long an

horsena [70]

Answer:

The average magnetic flux through each turn of the inner solenoid is $11.486\times10^{-8}\ Wb$

Explanation:

Given that,

Number of turns = 22 turns

Number of turns another coil = 330 turns

Length of solenoid = 21.0 cm

Diameter = 2.30 cm

Current in inner solenoid = 0.140 A

Rate = 1800 A/s

Suppose For this time, calculate the average magnetic flux through each turn of the inner solenoid

We need to calculate the magnetic flux

Using formula of magnetic flux

$\phi=BA$

$\phi=\dfrac{\mu_{0}N_{2}I}{l}\times\pi r^2$

Put the value into the formula

$\phi=\dfrac{4\pi\times10^{-7}\times330\times0.140}{21.0\times10^{-2}}\times\pi\times(\dfrac{2.30\times10^{-2}}{2})^2$

$\phi=11.486\times10^{-8}\ Wb$

Hence, The average magnetic flux through each turn of the inner solenoid is $11.486\times10^{-8}\ Wb$

7 0

3 years ago

A bottle lying on the windowsill falls off and takes 4.95 seconds to reach the ground. The distance from the windowsill to the g

Liula [17]

The distance an object falls from rest through gravity is
D = (1/2) (g) (t²)
   Distance = (1/2 acceleration of gravity) x (square of the falling time)

We want to see how the time will be affected
if ' D ' doesn't change but ' g ' does.
So I'm going to start by rearranging the equation
to solve for ' t '.                                                      D = (1/2) (g) (t²)

Multiply each side by 2 :         2 D =            g    t²

Divide each side by ' g ' :      2 D/g =                  t²

Square root each side:        t = √ (2D/g)

Looking at the equation now, we can see what happens to ' t ' when only ' g ' changes:

-- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'

   and smaller 'g' ==> longer 't' .--

They don't change by the same factor, because 1/g is inside the square root. So 't' changes the same amount as √1/g does.

Gravity on the surface of the moon is roughly 1/6 the value of gravity on the surface of the Earth.

So we expect ' t ' to increase by √6 = 2.45 times.

It would take the same bottle (2.45 x 4.95) = 12.12 seconds to roll off the same window sill and fall 120 meters down to the surface of the Moon.

5 0

3 years ago

What is the DISPLACEMENT of the motorbike rider in the picture?

lutik1710 [3]

120m north east hope this helps

6 0

3 years ago

A series RLC circuit has a resonant frequency = 6.00 kHz. When it is driven at a frequency = 8.00 kHz, it has an

ANEK [815]

The resistance (R) of the circuit is 707.1 ohms and the inductance (L) is 0.032 H.

<h3>Resistance of the circuit</h3>

For the phase constant of 45⁰, impedance is equal to the resistance of the circuit.

$Z= R\sqrt{2} \\\\R = \frac{Z}{\sqrt{2} } \\\\R = \frac{1000}{\sqrt{2} } = 707.1 \ ohms$

<h3>Resonant frequency</h3>

$f = \frac{1}{2\pi \sqrt{LC} } \\\\6000 = \frac{1}{2\pi \sqrt{LC} } \\\\2\pi(6000) = \frac{1}{\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi (6000)} \\\\LC = (\frac{1}{2\pi (6000)})^2\\\\LC = 7.034 \times 10^{-10} \\\\ C = \frac{7.034 \times 10^{-10} }{L} ---(1)$

<h3>At driven frequency</h3>

$X_l- X_c = R\\\\\omega L - \frac{1}{\omega C} = 707.1\\\\2\pi f L - \frac{1}{2\pi fC} = 707.1\\\\2\pi (8000) L - \frac{1}{2\pi (8000) C } = 707.1\ \ --(2)\\\\$

<em>solve 1 and 2 together</em>

$2\pi(8000) L - \frac{L}{2\pi (8000)(7.034 \times 10^{-10})} = 707.1\\\\50272L - 28279.48L = 707.1\\\\L = 0.032 \ H$

Learn more about impedance of RLC circuit here: brainly.com/question/372577

7 0

3 years ago

Write a definition for the word epicenter

docker41 [41]

Explanation: Epicenter is the part of the earth's surface directly above the focus of an earthquake.

I hope this helps you.

3 0

3 years ago