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.
<span><span>Area of light-blue triangle -
<span>The width of the triangle is 4 seconds and the height is 8 meters per second. To find the area, you use the equation: <span>area of triangle = 1⁄2 × base × height </span><span>so the area of the light-blue triangle is 1⁄2 × 8 × 4 = 16m. </span></span></span><span> 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.</span><span> Area under the whole graph
<span>The area of the light-blue triangle plus the area of the dark-blue rectangle is:16 + 48 = 64m.<span>This is the total area under the distance-time graph. This area represents the distance covered.</span></span></span></span>
Letter B
without a medium, there is nothing to compress, hence, no wave. A fast- medium like a gas (air) is easy to compress and allows waves to move through it easily. a slow medium, like a liquid, is still pretty fast, but not as fast as air.
Answer:
1 W = 1 J / sec Definition of watt is 1 joule / sec
So if a bulb uses 75 J / sec it must use
75 J/s * 60 sec / min = 4500 J/min energy used by bulb
If bulb is 15% efficient then the light delivered is
P = 4500 J / min * .15 = 675 J / min
Answer:
4.9 x 10^-19 J, 2.7 x 10^-19 J
Explanation:
first wavelength, λ1 = 410 nm = 410 x 10^-9 m
Second wavelength, λ2 = 750 nm = 750 x 10^-9 m
The relation between the energy and the wavelength is given by
E = h c / λ
Where, h is the Plank's constant and c be the velocity of light.
h = 6.63 x 10^-34 Js
c = 3 x 10^8 m/s
So, energy correspond to first wavelength
E1 = (6.63 x 10^-34 x 3 x 10^8) / (410 x 10^-9) = 4.85 x 10^-19 J
E1 = 4.9 x 10^-19 J
So, energy correspond to second wavelength
E2 = (6.63 x 10^-34 x 3 x 10^8) / (750 x 10^-9) = 2.652 x 10^-19 J
E2 = 2.7 x 10^-19 J
