Answer:0.00125 watts
Explanation:
resistance=50 ohms
Current=5 milliamps
Current=5/1000 milliamps
Current =0.005 amps
power=(current)^2 x (resistance)
Power=(0.005)^2 x 50
Power=0.005 x 0.005 x 50
Power=0.00125 watts
<h2>
Answer: Diffraction</h2><h2 />
Diffraction is a characteristic phenomenon that occurs in all types of waves
.
In this sense, <u>diffraction</u> happens when a wave (the light in this case) meets an obstacle or a slit .When this occurs, the light bends around the corners of the obstacle or passes through the opening of the slit that acts as an obstacle, forming <u><em>multiple patterns</em></u> with the shape of the aperture of the slit.
Note that the principal condition for the occurrence of this phenomena is that <u>the obstacle must be comparable in size (similar size) to the size of the wavelength.
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Decreased it because you can float a lot
Answer:

Explanation:
<u>Net Forces and Acceleration</u>
The second Newton's Law relates the net force
acting on an object of mass m with the acceleration a it gets. Both the net force and the acceleration are vector and have the same direction because they are proportional to each other.

According to the information given in the question, two forces are acting on the swimming student: One of 256 N pointing to the south and other to the west of 104 N. Since those forces are not aligned, we must add them like vectors as shown in the figure below.
The magnitude of the resulting force
is computed as the hypotenuse of a right triangle


The acceleration can be obtained from the formula

Note we are using only magnitudes here



<span>37.8 seconds
First, determine the speed difference between the car and truck.
95 km/h - 75 km/h = 20 km/h
Convert that speed into m/s to make a more convenient unit of measure.
20 km/h * 1000 m/km / 3600 s/h = 5.556 m/s
Now it's simply a matter of dividing the distance between the two vehicles and their relative speed.
210 m / 5.556 m/s = 37.8 s
So it will take 37.8 seconds for the car to catch the truck that's 210 meters in front of the car.</span>