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2 years ago

The following acceleration vs time plots show

Physics

1 answer:

fenix001 [56]2 years ago

7 0

The data set that is most likely from the car equipped with ABS is :

The First ( upper ) plot

<u><em>Although the plots related to your question is missing attached below is the missing data. </em></u>

The First plot ( acceleration vs time ) plot is the data that represents the application of brakes in a car fitted with ABS ( antilock braking system ) because of the absence of skidding effect between 6.5 and 9 seconds in the graph.

The skidding effect is present in the second plot around the same time because the car is not fitted with ABS

Hence we can conclude that the graph that shows the car fitted with ABS is the first plot .

Learn more : brainly.com/question/4360615

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Study the following reaction carefully. What classification should this reaction have?

Mariulka [41]

Answer:

Synthesis reaction

Explanation:

Synthesis reaction: two or more compounds combine to form one.

A + B → C

Decomposition reaction: one compound forms two or more.

C → A + B

Single replacement reaction: an element in one compound is replaced with another element.

AB + C → AC + B

Double replacement reaction: elements in two compounds replace each other.

AB + CD → AC + BD

6 0

3 years ago

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6th grade science !!!!!!!!!!!!!!!

posledela

Answer:

Explanation:

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3 years ago

A closely wound, circular coil with a diameter of 4.30 cm has 470 turns and carries a current of 0.460 A .

Nadusha1986 [10]

Hi there!

a)
Let's use Biot-Savart's law to derive an expression for the magnetic field produced by ONE loop.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}$

dB = Differential Magnetic field element

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

R = radius of loop (2.15 cm = 0.0215 m)

i = Current in loop (0.460 A)

For a circular coil, the radius vector and the differential length vector are ALWAYS perpendicular. So, for their cross-product, since sin(90) = 1, we can disregard it.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{r^2}$

Now, let's write the integral, replacing 'dl' with 'ds' for an arc length:
$B = \int \frac{\mu_0}{4\pi} \frac{ids}{R^2}$

Taking out constants from the integral:
$B =\frac{\mu_0 i}{4\pi R^2} \int ds$

Since we are integrating around an entire circle, we are integrating from 0 to 2π.

$B =\frac{\mu_0 i}{4\pi R^2} \int\limits^{2\pi R}_0 \, ds$

Evaluate:
$B =\frac{\mu_0 i}{4\pi R^2} (2\pi R- 0) = \frac{\mu_0 i}{2R}$

Plugging in our givens to solve for the magnetic field strength of one loop:

$B = \frac{(4\pi *10^{-7}) (0.460)}{2(0.0215)} = 1.3443 \mu T$

Multiply by the number of loops to find the total magnetic field:
$B_T = N B = 0.00631 = \boxed{6.318 mT}$

Now, we have an additional component of the magnetic field. Let's use Biot-Savart's Law again:
$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}$

In this case, we cannot disregard the cross-product. Using the angle between the differential length and radius vector 'θ' (in the diagram), we can represent the cross-product as cosθ. However, this would make integrating difficult. Using a right triangle, we can use the angle formed at the top 'φ', and represent this as sinφ.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} sin\theta}{r^2}$

Using the diagram, if 'z' is the point's height from the center:

$r = \sqrt{z^2 + R^2 }\\\\sin\phi = \frac{R}{\sqrt{z^2 + R^2}}$

Substituting this into our expression:
$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{(\sqrt{z^2 + R^2})^2} }(\frac{R}{\sqrt{z^2 + R^2}})\\\\dB = \frac{\mu_0}{4\pi} \frac{iRd\vec{l}}{(z^2 + R^2)^\frac{3}{2}} }$

Now, the only thing that isn't constant is the differential length (replace with ds). We will integrate along the entire circle again:
$B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} \int\limits^{2\pi R}_0, ds$

Evaluate:
$B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} (2\pi R)\\\\B = \frac{\mu_0 iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}$

Multiplying by the number of loops:
$B_T= \frac{\mu_0 N iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}$

Plug in the given values:
$B_T= \frac{(4\pi *10^{-7}) (470) (0.460)(0.0215)^2}{2 ((0.095)^2 + (0.0215)^2)^\frac{3}{2}}} \\\\ = 0.00006795 = \boxed{67.952 \mu T}$

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1 year ago

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A mountaintop is a height y above the level ground. A woman measures the angle of elevation of the mountaintop to be θ when she

valentina_108 [34]

Let t = Theta and p = Phi
Tan t = y/x Then x =y/Tant.
Tant = y/(x-d) x-d = y/Tanp
y/Tant - d = y/Tanp
y -d*Tanr = y*Tant/Tanp
y-y*Tant/Tanp = d*Tanr
y(1 - Tanr/Tanp = d*Tant

y = d*Tant/(1-Tant/Tanp)

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3 years ago

Which items do NOT describe possible solutions to air pollution?

bonufazy [111]

Answer:

2, 6

Explanation:

2 because if you cut down more trees you will have less items to help collect co2

6 because if you have more manufracturing more gasses will be release and moe carbon dioxcide in the air so it will slowly kill the ozone layer.

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3 years ago