The net force on particle particle q1 is 13.06 N towards the left.
<h3>
Force on q1 due to q2</h3>
F(12) = kq₁q₂/r₂
F(12) = (9 x 10⁹ x 13 x 10⁻⁶ x 7.7 x 10⁻⁶)/(0.25²)
F(12) = -14.41 N (towards left)
<h3>Force
on q1 due to q3</h3>
F(13) = (9 x 10⁹ x 7.7 x 10⁻⁶ x 5.9 x 10⁻⁶)/(0.55²)
F(13) = 1.352 N (towards right)
<h3>Net force on q1</h3>
F(net) = 1.352 N - 14.41 N
F(net) = -13.06 N
Thus, the net force on particle particle q1 is 13.06 N towards the left.
Learn more about force here: brainly.com/question/12970081
#SPJ1
We anticipate a constant Poynting vector of magnitude since the hot resistor will be emitting heat and none of the electric or magnetic fields will change over time.
S = P/A
= I2R/ 2πrL
= 332 kW/m2
Always pointing away from the wire, this Poynting vector.
<h3>What is the Poynting vector?</h3>
Describes the size and direction of the energy flow in electromagnetic waves using a Poynting vector. It bears the name of the 1884 invention of English physicist John Henry Poynting. It stands for the electromagnetic field's directional energy flux or power flow. The Poynting vector is significant in a static electromagnetic field because it determines the direction of energy flow in an electromagnetic field. This vector represents the radiation pressure of an electromagnetic wave and points in its direction of propagation.
To learn more about Poynting vector, visit:
<u>brainly.com/question/17330899</u>
#SPJ4
so the monster passige may be confusing so just do auto books
The problem states that the distance travelled (d) is
directly proportional to the square of time (t^2), therefore we can write this in
the form of:
d = k t^2
where k is the constant of proportionality in furlongs /
s^2
<span>Using the 1st condition where d = 2 furlongs, t
= 2 s, we calculate for the value of k:</span>
2 = k (2)^2
k = 2 / 4
k = 0.5 furlongs / s^2
The equation becomes:
d = 0.5 t^2
Now solving for d when t = 4:
d = 0.5 (4)^2
d = 0.5 * 16
<span>d = 8 furlongs</span>
<span>
</span>
<span>It traveled 8 furlongs for the first 4.0 seconds.</span>
