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2 years ago

Will the positioning of your Slinky along the east-west, north-south, or some other orientation affect your readings? Why?

1 answer:

5 0

Positioning your Slinky along any direction different from its initial position will affect your reading, because there will be change in the magnetic field.

<h3>Effect of magnet on Slinky</h3>

If the Slinky is made of an iron alloy, it can be magnetized by itself. Moving the Slinky around can cause a change in the magnetic field, even if no current is flowing.

When there is a change in the magnetic field, the reading changes.

At any point, you change the orientation of the Slinky, you will need to zero the reading or adjust the Slinky back to its initial position, even if the sensor does not move.

Thus, Positioning your Slinky along any direction that is different to its initial position will affect your reading because there will be change in the magnetic field.

Learn more about magnetic field here: brainly.com/question/7802337

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obesity occurs due to (a) overeating of carbohydrates and fats(b) not eating enough carbohydrates and fats (c) overeating of vit

aniked [119]

Answer:

B for BIG

Explanation:

carbs and fats is energy, if not being use or burned off, it'll be becomes body fat.

5 0

3 years ago

Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,

Liula [17]

Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

Explanation:

To solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

$F = \frac{k*q_1*q_2}{d^2}$ Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)

d: distance between the charges in meters (m)

Equivalence

1μC= 10⁻⁶C

1m = 100 cm

Data

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

$F_{13} = \frac{k*q_1*q_3}{d_1^2}$

$F_{23} = \frac{k*q_2*q_3}{d_2^2}$

F₁₃ = F₂₃

$\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2}$ We eliminate k and q₃ on both sides

$\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}$

$\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}$

$\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2}$ We eliminate 10⁻⁶ on both sides

$(1-x)^2 = \frac{7}{5} x^2$

$1-2x+x^2=\frac{7}{5} x^2$

$5-10x+5x^2=7 x^2$

$2x^2+10x-5=0$

We solve the quadratic equation:

$x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m$

$x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m$

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in opposite way .

x = 0.458m = 45.8cm

8 0

3 years ago

Your oven has a power rating of 5000 watt...

romanna [79]

A). 1,000 watts = 1 kilowatt
5,000 watts = 5 kilowatts

b). (5 kilowatts) x (2 hours) = 10 kilowatt-hours (kWh)

c). (10 kWh) x (15 cents/kWh) = $1.50

4 0

3 years ago

A velocity selector in a mass spectrometer uses an electric field of 4.4 x 105 V/m. What magnetic field strength (in Tesla) is n

Brut [27]

Answer:

B = 9.16 10⁻² T

Explanation:

The speed selector is a configuration where the electric and magnetic force has the opposite direction, which for a specific speed cancel

q v B = q E

v = E / B

B = E / v

Let's calculate

B = 4.4 10⁵ / 4.8 10⁶

B = 9.16 10⁻² T

5 0

3 years ago

Calculate the number of moles of water molecules in 12 dm' of water<br>vapour at STP.<br><br>

Vinvika [58]

Answer:

$22.4 \: {dm}^{3} \: are \: occupied \: by \: 1 \: mole \\ 12 \: {dm}^{3} \: will \: be \: occupied \: by \: ( \frac{12}{22.4} ) \: moles \\ = 0.536 \: moles$

4 0

3 years ago