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3 years ago

Which element could have the dot diagram shown below?

Physics

1 answer:

Lemur [1.5K]3 years ago

7 0

This answer is fluorine

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Your car's 30.0 W headlight and 2.50 kW starter are ordinarily connected in parallel in a 12.0 V system. What power (in W) would

Tatiana [17]

Answer:

<h3>The power of headlight in series connection is 29.64 W</h3>

Explanation:

Given :

Power of headlight $P_{1} = 30$ W

Power of starter $P_{2} = 2500$ W

Voltage of headlight and starter $V = 12$ V

From equation of power,

$P = \frac{V^{2} }{R}$

$R = \frac{V^{2} }{P}$

For finding the resistance of headlight and starter,

⇒ For headlight,

$R_{1} = \frac{144}{30} = 4.8$ Ω

⇒ For starter,

$R_{2} = \frac{144}{2500} = 0.057$ Ω

Since equivalent resistance,

$R_{eq} = R_{1} + R_{2} + ........$

$R_{eq} = 4.8 +0.057 = 4.857$ Ω

So power in series is given by,

$P_{s } = \frac{V^{2} }{R_{eq} } = \frac{144}{4.857}$

$P_{s } = 29.64$ W

8 0

3 years ago

Light shines through a single slit whose width is 5.7 x 10-4 m. A diffraction pattern is formed on a flat screen located 4.0 m a

lilavasa [31]

Answer:

$\lambda = 570\ nm$

Explanation:

Given,

Width of slit, W = 5.7 x 10⁻⁴ m

Distance between central bright fringe, L = 4 m

distance between central bright fringe and first dark fringe, y = 4 mm

Diffraction angle

$tan \theta = \dfrac{y}{L}$

$tan \theta = \dfrac{4}{4\times 10^3}$

$\theta = 0.0572$

Now.

$W sin \theta = m \lambda$

m = 1

$5.7 \times 10^{-4} \times sin (0.0572) = 1 \times \lambda$

$\lambda = 569.99 \times 10^{-9}\ m$

$\lambda = 570\ nm$

4 0

3 years ago

A series AC circuit contains a resistor, an inductor of 150 mH, a capacitor of 5.00 mF, and a generator with DVmax 5 240 V opera

yanalaym [24]

Given:

Inductance, L = 150 mH

Capacitance, C = 5.00 mF

$\Delta V_{max}$ = 240 V

frequency, f = 50Hz

$I_{max}$ = 100 mA

Solution:

To calculate the parameters of the given circuit series RLC circuit:

angular frequency, $\omega$ = $2\pi f = 2\pi \times50 = 100\pi$

a). Inductive reactance, $X_{L}$ is given by:

$\X_{L} = \omega L = 100\pi \times 150\times 10^{-3} = 47.12\Omega$

$X_{L} = 47.12\Omega$

b). The capacitive reactance, $X_{C}$ is given by:

$\X_{C} = \frac{1}{\omega C} = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 50\times 5.00\times 10^{-3}} = 0.636\Omega$

$X_{C} = 0.636\Omega$

c). Impedance, Z = $\frac{\Delta V_{max}}{I_{max}} = \frac{240}{100\times 10^{-3}} = 2400\Omega$

$Z = 2400\Omega$

d). Resistance, R is given by:

$Z = \sqrt {R^{2} + (X_{L} - X_{C})}$

$2400^{2} = R^{2} + (47.12 - 0.636)^{2}$

$R = \sqrt {5757839.238}$

$R = 2399.5\Omega$

e). Phase angle between current and the generator voltage is given by:

$tan\phi = \frac{X_{L} - X_{C}}{R}$

$\phi =tan^{-1}( \frac{X_{L} - X_{C}}{R})$

$\phi =tan^{-1}( \frac{47.12 - 0.636}{2399.5}) = tan^{-1}{0.0.01937}$

$\phi = 1.11^{\circ}$

5 0

3 years ago

You drop a 0.375 kg ball from a height of 1.37 m. It hits the ground and bounces up again to a height of 0.67 m. How much energy

Radda [10]

2.57 joule energy lose in the bounce .

<u>Explanation</u>:

when ball is the height of 1.37 m from the ground it has some gravitational potential energy with respect to hits the ground

Formula for gravitational potential energy given by

Potential Energy = mgh

Where ,

m = mass

g = acceleration due to gravity

h = height

Potential energy when ball hits the ground

m= 0.375 kg

h = 1.37 m

g = 9.8 m/s²

$Potential Energy = 0.375\times9.8\times1.37$

Potential Energy = 5.03 joule

Potential energy when ball bounces up again

h= 0.67 m

$Potential Energy = 0.375\times0.67\times9.8$

Potential Energy = 2.46 joule

Energy loss = 5.03 - 2.46 = 2.57 joule

2.57 joule energy lose in the bounce

6 0

3 years ago

How can you incorporate the information

il63 [147K]

Explanation:

There are many ways to achieve a healthier lifestyle and increase fitness. Healthy eating and constant exercise are essential for maintaining good physical and mental health. Through effective nutrition the human body is able to function better, there is an increase in disposition, improvement in immunity, improvement in blood rates, etc. Physical exercise, on the other hand, assists in strengthening muscles, controlling blood pressure, preventing cardiovascular disease, greater satisfaction, less stress, etc.

Some simple examples that can assist in improving health and fitness are:

Reduction of processed foods, sugars and soft drinks.

Increased fiber intake
Daily walks of at least 30 minutes
Increased water consumption
Increased consumption of fruits and vegetables
Incorporate meditation and yoga practices

6 0

2 years ago