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4 years ago

How to find the gradient of a velocity time graph

Physics

1 answer:

Sidana [21]4 years ago

7 0

The area-

The area under the line in a velocity-time graph represents the distance travelled. To find the distance travelled in the graph above, we need to find the area of the light-blue triangle and the dark-blue rectangle.

Area of light-blue triangle -
The width of the triangle is 4 seconds and the height is 8 meters per second. To find the area, you use the equation: area of triangle = 1⁄2 × base × height so the area of the light-blue triangle is 1⁄2 × 8 × 4 = 16m. Area of dark-blue rectangle The width of the rectangle is 6 seconds and the height is 8 meters per second. So the area is 8 × 6 = 48m. Area under the whole graph The area of the light-blue triangle plus the area of the dark-blue rectangle is:16 + 48 = 64m.This is the total area under the distance-time graph. This area represents the distance covered.

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What is the type of intermolecular force in methylated spirits

Olin [163]

. Methylated spirits have ethanol as a base but may include methyl alcohol (methanol) as part of the denaturing process.

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3 years ago

According to the theory of plate tectonics, the lithosphere is separated into sections that are called tectonic plates. These pl

loris [4]

Answer:

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3 years ago

A container is ﬁlled with water to a depth of 26.2 cm. On top of the water ﬂoats a 16.1 cm thick layer of oil with a density of

Solnce55 [7]

Answer:

1.34 x 10^3 Pa

Explanation:

density of oil = 0.85 x 10^3 kg/m^3

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Thus, the pressure on the surface of water is 1.34 x 10^3 Pa.

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3 years ago

Monochromatic light with a wavelength of 384 nm passes through a single slit and falls on a screen 86 cm away. If the distance o

valkas [14]

This is a Fraunhofer single slit experiment, where the light passing through the slit produces an interference pattern on the screen, and where the dark bands (minima of diffraction) are located at a distance of
$y= \frac{m\lambda D}{a}$
from the center of the pattern. In the formula, m is the order of the minimum, $\lambda$ the wavelenght, $D$ the distance of the screen from the slit and $a$ the width of the slit.

In our problem, the distance of the first-order band (m=1) is $y=0.22 cm$ . The distance of the screen is D=86 cm while the wavelength is $\lambda = 384 nm=384 \cdot 10^{-7}cm$ . Using these data and re-arranging the formula, we can find a, the width of the slit:
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3 years ago

the half-life of carbon-14 is 5,730 years. After 11,460 year, how much of original carbon-14 remains?

Inessa [10]

Via the half-life equation:

$A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h}}$

Where the time elapse is 11,460 year and the half-life is 5,730 years.

$A_{final}=A_{initial}(\frac{1}{2})^\frac{11460}{5730} \\\\A_{final}=A_{initial}\frac{1}{4} \\\\A_{final}=\frac{1}{4}A_{initial}$

Therefore after 11,460 years the amount of carbon-14 is one fourth (1/4) of the original amount.

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3 years ago