The magnitude of the electrical force between a pair of charged particles is 4 Times as much when the particles are moved half as far apart.
This can be easily understood by Columb's law,
which state's that the amount of electrical force experienced by two charged particles is inversely proportional to the square of the distance between them.
∴
Now, we know the new distance is half the original distance,
The electrical force of attraction or electrostatic force of attraction between two charged particles refers to the amount of attractive or repulsive force that exists between the two charges. This can be calculated by Columb's Law.
A charged particle in physics is a particle that has an electric charge. It might be an ion, such as a molecule or atom having an excess or shortage of electrons in comparison to protons. The same charge is thought to be shared by an electron, a proton, or another primary particle.
Learn more about electrical force here
brainly.com/question/2526815
#SPJ4
The time taken for the mass to reach the bottom of the inclined plane after it is released from rest is 0.78 s.
<h3>
Height of the inclined plane</h3>
The height of the inclined plane is calculated as follows;
h = L(sin 37)
where;
- h is height of the plane
- L is length of the plane
h = 5 x sin(37)
h = 3.01 m
<h3>Time of motion of the mass</h3>
t = √(2h/g)
t = √(2 x 3.01 / 9.8)
t = 0.78 s
Thus, the time taken for the mass to reach the bottom of the inclined plane after it is released from rest is 0.78 s.
Learn more about time of motion here: brainly.com/question/2364404
#SPJ1
Answer:
Option D is correct: 170 µW/m²
Explanation:
Given that,
Frequency f = 800kHz
Distance d = 2.7km = 2700m
Electric field Eo = 0.36V/m
Intensity of radio signal
The intensity of radial signal is given as
I = c•εo•Eo²/2
Where c is speed of light
c = 3×10^8m/s
εo = 8.85 × 10^-12 C²/Nm²
I = 3×10^8 × 8.85×10^-12 × 0.36²/2
I = 1.72 × 10^-4W/m²
I = 172 × 10^-6 W/m²
I = 172 µW/m²
Then, the intensity of the radio wave at that point is approximately 170 µW/m²
Answer:
A.
Explanation:
NEAR THE CENTER OF TECTONIC PLATES.
Answer:
7650 m.
Explanation:
Ocean floor depth, d is:
d = v * t,
where,
d = the distance from the vessel to the ocean floor (or the depth)
v = 1530 m/s = velocity of the ultrasonic sound
t = t_echo/2 = time that the ultrasonic sound needs to reach the ocean floor
t_echo = 10 s = time that the ultrasonic sound needs to reach the ocean floor and return back to the vessel.
d = v * t
= v * t_echo/2
= 1530 * 10/2
= 7650 m.