Answer:
The magnitude of the electric field at a point equidistant from the lines is 
Explanation:
Given that,
Positive charge = 24.00 μC/m
Distance = 4.10 m
We need to calculate the angle
Using formula of angle
We need to calculate the magnitude of the electric field at a point equidistant from the lines
Using formula of electric field
Put the value into the formula
Hence, The magnitude of the electric field at a point equidistant from the lines is
The only thing you need to know in order to solve this task is that <span>plank length (which is force x), should equal the increase in potential energy, so what we have now : (mass)* g * (height).
It has to look like that: </span>
<span>F * 3.0 = 150 x 9.81 x 1.20
Then solve for F, the result should be in newtones = 588N
Do hope it makes sense.</span>
Answer:
<h2>17.1 N</h2>
Explanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question
3800 g = 3.8 kg
We have
force = 3.8 × 4.5
We have the final answer as
<h3>17.1 N</h3>
Hope this helps you
Our bodies emit heat, and nerve endings in our skin can detect it.
Our eyes can detect visible light, but our bodies don't emit that.
Answer:
C = 17 i^ - 7 j^ + 16 k^
, | C| = 24.37
Explanation:
To work the vactor component method, we add the sum in each axis
C = A + B = (Aₓ + Bₓ) i ^ + (
+ 
) i ^ + (
+ 
) k ^
Cₓ = 12+ 5 = 17
= -37 +30 = -7
= 58 -42 = 16
Resulting vector
C = 17 i ^ - 7j ^ + 16k ^
The mangitude of the vector is
| C | = √ c²
| C | = √( 17² + 7² + 16²)
| C| = 24.37
