The task is to show that the right side of the equation has units of [Time], just like the left side has.
The right side of the equation is . . . 2 π √(L/G) .
We can completely ignore the 2π since it has no units at all, so it has no effect on the units of the right side of the equation. Now the task is simply to find the units of √(L/G) .
L . . . meters
G . . . meters/sec²
(L/G) = (meters) / (meters/sec²)
(L/G) = (meters) · (sec²/meters)
(L/G) = (meters · sec²) / (meters)
(L/G) = sec²
So √(L/G) = seconds = [Time]
THAT's what we were hoping to prove, and we did it !
Answer:
https://www.chegg.com/homework-help/questions-and-answers/part-20-cm-long-stick-m-0400-kg-lifted-rope-tied-70-cm-upper-end-end-touches-smooth-floor--q81268572
Answer:
the rate of the change of the length of the shadow is - 0.8625 m/s.
The negative(-) sign means the length of the shadow decreases at a rate of 0.8625 m/s.
Explanation:
Given the data in the question;
Let x represent the man's distance from building,
initially x = 1m2
dx/d t= -2.3 m/s
Also Let y represent shadow height
so we determine dy/dt when x is 4m from the building
form the image description of the problem, we see two-like triangles with the same base and height ratios
so
2 / (12-x) = y / 12
24 = y(12 - x )
y = 24 / (12-x)
dy/dt = 24/(12-x)² × dx/dt
Now at x = 4,
we substitute
dy/dt will be;
⇒ 24/(12 - 4)² × -2.3
= 24/64 - 2.3
= 0.375 × -2.3
dy/dt = - 0.8625 m/s
Therefore, the rate of the change of the length of the shadow is - 0.8625 m/s.
The negative(-) sign means the length of the shadow decreases at a rate of 0.8625 m/s.
Answer:
179.47m/s
Explanation:
Using the law of conservation of momentum
m1u1 + m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and u2 are the initial velocities
v is the final velocity
Substitute
7750(179)+72(230) = (7750+72)v
1,387,250+16560 = 7822v
1,403,810 = 7822v
v = 1,403,810/7822
v= 179.47m/s
Hence the final velocity of the probe is 179.47m/s
