His total displacement from his original position is -1 m
We know that total displacement of an object from a position x to a position x', d = final position - initial position.
d = x' - x
If we assume the lad's initial position in front of her house is x = 0 m. The lad then moves towards the positive x-axis, 5 m. He then ends up at x' = 5 m. He then finally goes back 6 m.
Since displacement = final position - initial position, and his displacement is d' = -6 m (since he moves in the negative x - direction or moves back) from his initial position of x' = 5 m.
His final position, x" after moving back 6 m is gotten from
x" - x' = -6 m
x" = -6 + x'
x" = -6 + 5
x" = -1 m
Thus, his total displacement from his original position is
d = final position - initial position
d = x" - x
d = -1 m - 0 m
d = -1 m
So, his total displacement from his original position is -1 m
Learn more about displacement here:
brainly.com/question/17587058
Answer: Real image
Explanation:
converging lens will only produce a real image if the object is located beyond the focal point (i.e., more than one focal length away).
Answer:
(A) -2940 J
(B) 392 J
(C) 212.33 N
Explanation:
mass of bear (m) = 25 kg
height of the pole (h) = 12 m
speed (v) = 5.6 m/s
acceleration due to gravity (g) = 9.8 m/s
(A) change in gravitational potential energy (ΔU) = mg(height at the bottom- height at the top)
height at the bottom = 0
= 25 x 9.8 x (0-12) = -2940 J
(B) kinetic energy of the Bear (KE) = 
= 
= 392 J
(C) average frictional force = 
- change in KE (ΔKE) = initial KE - final KE
- ΔKE = -
- when the Bear reaches the bottom of the pole, the final velocity (Vf) is 0, therefore the change in kinetic energy becomes ΔKE =
- 0 = 392 J
\frac{-(ΔKE+ΔU)}{h}[/tex] = 
= 
= 212.33 N
Answer:
70.5 mph
Explanation:
A passenger jet travels from Los Angeles to Bombay, India, in 22h.
The return flight takes 17 h.
The difference in flight times is caused by winds over the Pacific Ocean that
blow primarily from west to east.
If the jet's average speed in still air is 550 mi/h what is the average speed
of the wind during the round trip flight? Round to the nearest mile per hour.
Is your answer reasonable?
:
Let w = speed of the wind
:
Write a distance equation (dist is the same both ways
17(550+w) = 22(550-w)
9350 + 17w = 12100 - 22w
17w + 22w = 12100 - 9350
39w = 2750
W = 2750/39
w = 70.5 mph seems very reasonable
:
Confirming if the solution by finding the distances using these value
17(550+70.5) = 10549 mi
22(550-70.5) = 10549 mi; confirms our solution of w = 70.5 mph
