D A little longer than a yard (about 3 ft)
Answer:
0.35 m
Explanation:
Constructive interference occurs when the difference in the distance between the two speakers is
0, λ, 2λ,...
Here λ = v/f = 343/490 = 0.7m
The first point of interference is the initial point which would mean the distance is the trivial solution of 0 .
A microphone is placed half-way between the speakers and then moved along the line joining the two speakers until the first point of constructive interference is found.
Let x be the distance from the midpoint to the next interference.
This occurs when 1.5 + x -(1.5 - x) = 0.7
so 2x = 0.7 which means x = 0.7/2 = 0.35m
Answer: 4.27 *10^6 N/C
Explanation: In order to calculate the electric field along the axis of charged ring we have to use the following expression:
E=k*x/(a^2+x^2)^3/2 where a is the ring radius and x the distance to the point measured from the center of the ring.
Replacing the data we have:
E= (9* 10^9* 0.3* 50 * 10^-6)/(0.1^2+0.3^2)^3/2
then
E=4.27 * 10^6 N/C
A complex entity involving the Earth's biosphere, atmosphere, oceans, and soil; the totality constituting a feedback or cybernetic system which seeks an optimal physical and chemical environment for life on this planet
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>
