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Ivahew [28]
4 years ago
6

Which of the following is true about scientific models? (2 points) Select one: a. Models are used to simplify the study of thing

s. b. Computer models are the most reliable kind of model. c. Models explain past, present and future information. d. A model is accurate if it does not change over time.
Physics
2 answers:
PIT_PIT [208]4 years ago
6 0

I believe the answer is A

Vanyuwa [196]4 years ago
5 0

Question: Which of the following is true about scientific models?  Select one: A) Models are used to simplify the study of things. B) Computer models are the most reliable kind of model. C) Models explain past, present and future information. D) A model is accurate if it does not change over time.

Answer: <u>A) Models are used to simplify the study of things.</u>

<em>Hope this helps!.</em>

<em>~~~~~~~~~~~~~~~~~~~~~~</em>

<em>~A.W~ZoomZoom44</em>

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Explanation:

4 0
3 years ago
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A 10.0 V battery is connected across two resistors in series. One resistor has resistance of 840.0 Ω and the other has resistanc
REY [17]

Answer:

Explanation:

There are a couple of ways you could do this.

The easiest is to use E*R1/(R1 + R2)

  • E = 10 volts
  • R1 = 590 ohms
  • R2 = 840 ohms

So the result would be

E_590 = 10 * 590/(590 + 840)

E_590 = 10 * 590/ (1430)

E_590 = 4.13 volts rounded.

You could do this a slightly longer way.

R = 1430 (total ohms in series.

E = 10 volts

I = ???

I = E/R

I = 10 / 1430

I = 0.00699

Now use this current to figure out the voltage drop.

E = I * R

I = 0.00699 amps

R = 590 ohms

E = 0.00699 * 590

E = 4.13 volts

Pick the way of doing it you like best.

4 0
3 years ago
Consider a coaxial cable (like the kind that is used to carry a signal to your TV). In this cable, a current I runs in one direc
Ira Lisetskai [31]

Answer:

0 < r < r_exterior     B_total = \frac{\mu_o I}{2\pi  r}

r > r_exterior            B_total = 0

Explanation:

The magnetic field created by the wire can be found using Ampere's law

        ∫ B. ds = μ₀ I

bold indicates vectors and the current is inside the selected path

           

outside the inner cable

          B₁ (2π r) = μ₀ I

          B₁ = \frac{\mu_o I}{2\pi  r}

the direction of this field is found by placing the thumb in the direction of the current and the other fingers closed the direction of the magnetic field which is circular in this case.

For the outer shell

for the case   r> r_exterior

         

           B₂ = \frac{\mu_o I}{2\pi  r}

This current is in the opposite direction to the current in wire 1, so the magnetic field has a rotation in the opposite direction

for the case r <r_exterior

in this case all the current is outside the point of interest, consequently not as there is no internal current, the field produced is zero

           B₂ = 0

Now we can find the field created by each part

0 < r < r_exterior

          B_total = B₁

          B_total = \frac{\mu_o I}{2\pi  r}  

r > r_exterior

          B_total = B₁ -B₂

          B_total = 0

6 0
3 years ago
Which one of the following experiments is the justification for the wave theory of light? a. Newton's rings experiment b. Frank-
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Answer:

c

Explanation:

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3 years ago
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Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance
scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

x_{1}=\dfrac{\lambda D}{2d}

Second minima from center

x_{2}=\dfrac{3\lambda D}{2d}

The distance between the places where the intensity is zero due to the double slit effect

\Delta x_{d}=x_{2}-x_{1}

\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}

\Delta x_{d}=\dfrac{\lambda D}{d}

Put the value into the formula

\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}

\Delta x_{d}=0.015 =15\times10^{-3}\ m

\Delta x_{d}=15\ mm

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

8 0
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