Answer: magnitude of resultant = 2.811km, direction of resultant = 209.6°
Explanation: The vector ( in this case displacement) that lies on the y axis is 3.25km north and 1.50km due south.
Their resultant is gotten below
Magnitude of resultant = 3.25 - 1.50 = 1.75km
Direction of resultant = north (direction of the bigger vector)
2.20km is the only vector acting on the x axis due east, combining this vector with the resultant of the vectors above, we realize that 2.2km west is perpendicular to 1.75km due north. Since 1.75km us due north, it implies that it is the vector on the positive y axis (vy) and 2.20km due west implies that it is the vector on the negative x axis(vx), thus their resultant is gotten using phythagoras theorem.
R = √ vx² + vy²
R = √ 2.20² + 1.75²
R = √ 7.9025
R= 2.1811km.
θ = tan^-1 (vy/vx)
θ = tan ^-1 (1.25/2.20)
θ = tan ^-1 (0.5618)
θ = 29.6°.
The direction of the vector is south west which implies the third quadrant of the trigonometric quadrant which implies 180 + θ
Thus the direction of the vector is 180 + 29.60 = 209.6°
From the picture attached to this answer we can see roughly that the magnitude of resultant is longer than it component and the vector is placed on the 3rd quadrant which verifies our quantitative claim.
Answer:
The value is
Explanation:
From the question we are told that
The maximized electric field strength is 
The diameter is 
Generally the cross -sectional area is mathematically represented as
=> 
=> 
=> 
Generally the maximum power is mathematically represented as
Here c is the speed of light with value 
is the permeability of free space with value 
=>
Answer:
v = 26.7 mph
Explanation:
During the first 5 hours, at a constant speed of 20 mph, we find the total displacement to be as follows:
Δx₁ = v₁*t₁ = 20 mph*5 h = 100 mi
Assuming we can neglect the displacement during the speeding up from 20 to 60 mph, we can find the the total displacement at 60 mph as follows:
Δx₂ = v₂*t₂ = 60 mph*1 h = 60 mi
So, the total displacement during all the trip wil be:
Δx = Δx₁ + Δx₂ = 100 mi + 60 mi = 160 mi
So we can find the the average velocity during the 6-hour period, applying the definition of average velocity, as follows:
v = Δx / Δt = 160 mi / 6 h = 26.7 mph