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hjlf
2 years ago
8

By constructing basic electric circuits, you will be able to measure current flow. With Ohm’s law in mind, what broad question a

re you answering by doing this experiment?
Physics
2 answers:
NemiM [27]2 years ago
8 0

Answer:

Sample Response: How do changes in voltage or resistance affect current in an electric circuit?

Explanation:

andrezito [222]2 years ago
5 0

Broad questions that we can answer by doing this experiment are <em><u>the effects of electric current</u></em>

<h3>Further explanation</h3>

Electric current is the amount of electric charge that flows each unit of time

Electric current occurs due to the movement of electrons due to the difference in potential or voltage (from high potential to low potential) between two points

Electrons will flow through the conducting wire that functions as a conductor

Ohm's Law states that

<em>The potential difference is proportional to the electric current as long as the resistance is constant</em>

\displaystyle I=\frac{V}{R}\\\\V=I\times R

A simple electrical circuit consists of a voltage source (battery) and a lamp

Ampermeters to measure the strength of the current, must be installed in series with the load to be measured

By changing the voltage source, with constant resistance from a conductor, different current measurements will be obtained. The greater the voltage, the greater the resulting current.

<h3>Learn more</h3>

electrons flow through the device

brainly.com/question/4438943

Keywords : basic electric circuits, Ohm’s law, experiment

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BlackZzzverrR [31]
Decomposing the vector b on the x-axis and the y-axis, we get a rectangle triangle where the two sides are bx (x-axis) and by (y-axis), and b is the hypothenuse.
The component in x, bx, is equal to the product between the hypothenuse and the cosine of the angle between b and the x-axis, which is 23.5 ^{\circ}:
b_x = b \cos (23^{\circ})=(4.00 m)(\cos (23^{\circ}))=3.68 m
6 0
2 years ago
luggage handler pulls a 20.0 kg suitcase up a ramp inclined at 34.0 ∘ above the horizontal by a force F⃗ of magnitude 165 N that
Andreas93 [3]

Answer:

a)  W = 643.5 J, b) W = -427.4 J  

Explanation:

a) Work is defined by

       W = F. x = F x cos θ

in this case they ask us for the work done by the external force F = 165 N parallel to the ramp, therefore the angle between this force and the displacement is zero

       W = F x

let's calculate

       W = 165  3.9

        W = 643.5 J

b) the work of the gravitational force, which is the weight of the body, in ramp problems the coordinate system is one axis parallel to the plane and the other perpendicular, let's use trigonometry to decompose the weight in these two axes

       sin θ = Wₓ / W

       cos θ = Wy / W

        Wₓ = W sinθ = mg sin θ

        Wy = W cos θ

the work carried out by each of these components is even Wₓ, it has to be antiparallel to the displacement, so the angle is zero

      W = Wₓ x cos 180

      W = - mg sin 34  x

     

let's calculate

       W = -20 9.8 sin 34 3.9

        W = -427.4 J

The work done by the component perpendicular to the plane is ero because the angle between the displacement and the weight component is 90º, so the cosine is zero.

3 0
2 years ago
Chromatic aberration comes from the fact that different wavelengths of light travel at different speeds through the material of
gtnhenbr [62]

Answer:

 y_red / y_blue = 1.11

Explanation:

Let's use the constructor equation to find the image for each wavelength

         1 /f = 1 /o + 1 /i

Where f is the focal length, or the distance to the object and i the distance to the image

Red light

           1 / i = 1 / f - 1 / o

           1 / i_red = 1 / f_red - 1 / o

           1 / i_red = 1 / 19.57 - 1/30

           1 / i_red = 1,776 10-2

           i_red = 56.29 cm

Blue light

            1 / i_blue = 1 / f_blue - 1 / o

            1 / i_blue = 1 / 18.87 - 1/30

            1 / i_blue = 1,966 10-2

            i_blue = 50.863 cm

Now let's use the magnification ratio

             m = y ’/ h = - i / o

             y ’= - h i / o

Red Light

            y_red ’= - 5 56.29 / 30

            y_red ’= - 9.3816 cm

Light blue

            y_blue ’= 5 50,863 / 30

            y_blue ’= - 8.47716 cm

The ratio of the height of the two images is

            y_red ’/ y_blue’ = 9.3816 / 8.47716

            y_red / y_blue = 1,107

            y_red / y_blue = 1.11

5 0
2 years ago
Calculate the heat required when 2.50 mol of a reacts with excess b and a2b according to the reaction: 2a + b + a2b → 2ab + a2 g
Yuri [45]

Answer:

Q = 12.5 kJ

Explanation:

The expression to use to calculate Heat is:

Q = H° * n

Where:

Q: heat (J or kJ)

H°: enthalpy of reaction (kJ/mol)

n: moles

Now, as it was stated in the comments, the question is incomplete, and here is the missing part:

Given:

2A + B  A2B (1)

H° = – 25.0 kJ/mol

2A2B  2AB + A2 (2)

H° = 35.0 kJ/mol

With these two reactions, we can calculate the heat.

Now, with the above two reactions, we need to get the general reaction (The one the question is giving), so, let's use (1) and (2) and do the sum of them:

2A + B -------> A2B   H°1 = -25 kJ/mol

2A2B --------> 2AB + A2   H°2 = 35 kJ/mol

Now, we sum both equations, we can see that one A2B cancels out with one A2B from equation 2, so, the equation gives:

2A + B + 2A2B -------> 2AB + A2

And the enthalpy, it's just summed:

H°3 = -25 + 35 = 10 kJ/mol

Now with this value we can calculate heat:

Q = 10 * 2.5 = 25 kJ

However, in the reaction we have 2A, so it's not 1:1 mole ratio, but instead is 1:2, so this result we have to divide it between 2 so:

<u>Q = 25 / 2 = 12.5 kJ</u>

3 0
2 years ago
A constant braking force of 10 newtons applied for 5 seconds is used to stop a 2.5-kilogram cart traveling at 20 meters per seco
ss7ja [257]
Mathematically it can be expressed as: I = FpΔt. Where F p is the average magnitude of the acting force and Δt = t 2 - t 1, the time lapse in the force acts.
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 I = FpΔt = (10N) * (5s) = 50 N.s
6 0
2 years ago
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