Imagine a ball is moving on the following horizontal line.
. . . . . . . . . . . . . . . . . . . O. . . . . . . . . . . . . . . . . .
Take right as positive. O is the starting point of the ball. Denote the ball by o.
. . . . . . . . . . . . . . . . . . . O. . . . . . . ... . . o . . . . . .
Assume the ball is moving to the right. It has positive displacement since it is on the right of O, and positive velocity since its positive displacement is increasing.
.ñ
. . . . . . . . . . . . . . . . . . . O. . . . o . . . . . . . . . . . . .
Now the ball is returning to O. It still has positive displacement since its current position is still on the right of O. However, its velocity is negative since its positive displacement is decreasing and the direction of the velocity vector points left, which is the negative side.
By now you should be able to come up with a scenario where the ball has negative displacement and positive velocity.
You can observe the same phenomenon in daily life. Say, as a stretched spring bounces to its starting position, if we let the returning direction be positive, the string has negative displacement since it is on the negative direction, but has positive velocity. Bungee jump can also used to illustrate the phenomenon.
The final velocity of the truck is found as 146.969 m/s.
Explanation:
As it is stated that the lorry was in standstill position before travelling a distance or covering a distance of 3600 m, the initial velocity is considered as zero. Then, it is stated that the lorry travels with constant acceleration. So we can use the equations of motion to determine the final velocity of the lorry when it reaches 3600 m distance.
Thus, a initial velocity (u) = 0, acceleration a = 3 m/s² and the displacement s is 3600 m. The third equation of motion should be used to determine the final velocity as below.
Then, the final velocity will be
Thus, the final velocity of the truck is found as 146.969 m/s.
Answer:
9.01amp
Explanation:
Power = V^2/R
Given that v = 11volts, P = 99watts
99 = 11^2/R
11×11 = 99R
121= 99R
R = 121/99
R= 1.22ohms
From ohms Law; V = IR
11volts = I × 1.22ohms
I = 11/1.23
I = 9.01 amp
Answer: The field lines bend away from the second positive charge
Explanation: opposite attracts, same repulse
Answer:
Intensity of the light (first polarizer) (I₁) = 425 W/m²
Intensity of the light (second polarizer) (I₂) = 75.905 W/m²
Explanation:
Given:
Unpolarized light of intensity (I₀) = 950 W/m²
θ = 65°
Find:
a. Intensity of the light (first polarizer)
b. Intensity of the light (second polarizer)
Computation:
a. Intensity of the light (first polarizer)
Intensity of the light (first polarizer) (I₁) = I₀ / 2
Intensity of the light (first polarizer) (I₁) = 950 / 2
Intensity of the light (first polarizer) (I₁) = 425 W/m²
b. Intensity of the light (second polarizer)
Intensity of the light (second polarizer) (I₂) = (I₁)cos²θ
Intensity of the light (second polarizer) (I₂) = (425)(0.1786)
Intensity of the light (second polarizer) (I₂) = 75.905 W/m²
