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3 years ago

Samuel places a ladder against his house. The base of the

Physics

2 answers:

NemiM [27]3 years ago

5 0

Answer:

d. 8 feet

Explanation:

(height)^2 = (10)^2 - (6)^2

= 100 - 36

height = √64

= 8 feet

yanalaym [24]3 years ago

4 0

Answer:

d. 8 feet

Explanation:

let :

a = length of the ladder = 10 feet

b = how high above the ground the ladder touches the wall

c = the length of the base of the ladder from the wall . 6 feet

By Pythagoras' theorem

${a }^{2} = {b}^{2} + {c}^{2}$

Therefore
${b}^{2} = {a}^{2} - {c}^{2}$
${b}^{2} = {10}^{2} - {6}^{2}$

${b}^{2} = 100 - 36$

${b}^{2} = 64$
Square root both sides
$\sqrt{ {b}^{2} } = \sqrt{64}$
$b = 8$
Therefore 8 feet above the ground the ladder touches the wall

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White raven [17]

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3 years ago

In a physics laboratory experiment, a coil with 250 turns enclosing an area of 10.5 cm2 is rotated during the time interval 3.10

olchik [2.2K]

I guess the problem is asking for the induced emf in the coil.

Faraday-Neumann-Lenz states that the induced emf in a coil is given by:
$\epsilon = -N \frac{\Delta \Phi}{\Delta t}$
where
N is the number of turns in the coil
$\Delta \Phi$ is the variation of magnetic flux through the coil
$\Delta t$ is the time interval

The coil is initially perpendicular to the Earth's magnetic field, so the initial flux through it is given by the product between the magnetic field strength and the area of the coil:
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At the end of the time interval, the coil is parallel to the field, so the final flux is zero:
$\Phi_f = 0$

Therefore, we can calculate now the induced emf by using the first formula:
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4 years ago

A circuit consists of one 330 ohm resistor in series with two other resistors (470 ohms and 220 ohms) connected in parallel. Wha

umka21 [38]

Answer:

power drain on an ideal battery, P = 0.017 W

Given:

$R_{1} = 330\ohm$

$R_{2} = 470\ohm$

$R_{3} = 220\ohm$

Since, $R_{2} = 470\ohm$ and $R_{3} = 220\ohm$ are in parallel and this combination is in series with $R_{1} = 330\ohm$ , so,

Equivalent resistance of the circuit is given by:

$R_{eq} = \frac{R_{2}R_{3}}{R_{2} + R_{3}} + R_{1}$

$R_{eq} = \frac{470\times 220}{470 + 220} + 330$

$R_{eq} = 149.85 + 330 = 479.85 \ohm$

power drain on an ideal battery, P = $\frac{V^{2}}{R_{eq}}$

P = $\frac{2.9^{2}}{479.85}$

P = 0.017 W

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4 years ago

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Mashcka [7]

Answer:

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3 years ago

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Answer:

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