South is perpendicular to West, so the plane's route forms a right triangle, and you can use Good Old Pythagoras to calculate the length of the hypotenuse.
The length of the displacement is √(the west piece² + the south piece²) .
That's √ [ (340km)² + (360km)² ]
= √ [ (115,600) km² + (129,600) km² ]
= √ 245,200 km²
= 495 km
To be technical, Displacement is a vector, so we would need to
calculate its direction too.
Naturally, the plane winds up roughly southwest of where it took off.
You'd want to find the angle whose tangent is (360/340) = about 1.059 .
The direction of the Displacement is that angle south of west. (about 46.6 degrees)
Answer:
200 J
Explanation:
Given that,
Force, F = 20 N
Distance covered in the direction of force, d = 10 m
We need to find the effort made by Mrs. Lasarati. It will be given by :
W = Fd
W = 20 N × 10 m
W = 200 J
So, the required effort made by Mrs. Lasarati is 200 J.
Answer:
0.21%
Explanation:
We are given;
Mass; m = 100 kg
Diameter; d = 2.2 mm = 2.2 × 10^(-3) m
Young's modulus; E = 12.5 x 10^(10) N/m².
Formula for area is;
A = πd²/4
A = (π/4) x (2.2 x 10^(-3))²
A = 3.8 x 10^(-6) m²
Force; F = mg
g is acceleration due to gravity and has a constant value of 9.8 m/s²
F = 100 × 9.8
F = 980 N
Formula for young's modulus is;
E = Stress/strain
Formula for stress = F/A
Formula for strain = ΔL/L
Thus;
E = (F/A)/(ΔL/L)
Making ΔL/L the subject, we have;
ΔL/L = (F/A)/E
Plugging in the relevant values;
ΔL/L = 980/(3.8 x 10^(-6) × 12.5 × 10^(10))
ΔL/L = 0.0021
Then percentage increase in length of a wire = 0.0021 × 100% = 0.21%
Answer:
C)The side that the charge is on, the top.
Explanation:
As we know that
1)More flux is possible, when the field is perpendicular to the cross sectional area.
2)Electrical field lines are more power full in this side because the charge is near to the top side.
3)Those sides are electric field parallel to area that sides have minimum flux.
So
The side that the charge is on, the top ,have least flux traverse.
Option C is correct.