suppose you want to determine the surface area of this sugar cube. it has edges that are each 2 cm long. if you cut the cube in
1 answer:
You might be interested in
Answer:
a) the magnitude of r is 184.62
b) the direction is 37.74° south of the negative x-axis
Explanation:
Given the data in the question;
as illustrated in the image blow;
To find the the magnitude of r, we will use the Pythagoras theorem
r² = y² + x²
r = √( y² + x²)
we substitute
r = √((-113)² + (-146)²)
r = √(12769 + 21316 )
r = √(34085 )
r = 184.62
Therefore, the magnitude of r is 184.62
To find its direction, we need to find ∅
from SOH CAH TOA
tan = opposite / adjacent
tan∅ = -113 / -146
tan∅ = 0.77397
∅ = tan⁻¹( 0.77397 )
∅ = 37.74°
Therefore, the direction is 37.74° south of the negative x-axis
Answer:
direction is Horizontal
Explanation:
As we know that the string is horizontal here
so the tension force in the string is due to electrostatic force on it
now we will have
so here the force is tension force on it
now we have
direction is Horizontal
1) Potential difference: 1 V
2)
Explanation:
1)
When a charge moves in an electric field, its electric potential energy is entirely converted into kinetic energy; this change in electric potential energy is given by
where
q is the charge's magnitude
is the potential difference between the initial and final position
In this problem, we have:
is the magnitude of the charge
is the change in kinetic energy of the particle
Therefore, the potential difference (in magnitude) is
2)
Here we have to evaluate the direction of motion of the particle.
We have the following informations:
- The electric potential increases in the +x direction
- The particle is positively charged and moves from point a to b
Since the particle is positively charged, it means that it is moving from higher potential to lower potential (because a positive charge follows the direction of the electric field, so it moves away from the source of the field)
This means that the final position b of the charge is at lower potential than the initial position a; therefore, the potential difference must be negative: