The answer is D. If you aren't consistent with your drop positions, then your data may be invalid. To be frank: it basically screws over the experiment.
True.
I think that’s the answer.
Answer:
12.0 meters
Explanation:
Given:
v₀ = 0 m/s
a₁ = 0.281 m/s²
t₁ = 5.44 s
a₂ = 1.43 m/s²
t₂ = 2.42 s
Find: x
First, find the velocity reached at the end of the first acceleration.
v = at + v₀
v = (0.281 m/s²) (5.44 s) + 0 m/s
v = 1.53 m/s
Next, find the position reached at the end of the first acceleration.
x = x₀ + v₀ t + ½ at²
x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²
x = 4.16 m
Finally, find the position reached at the end of the second acceleration.
x = x₀ + v₀ t + ½ at²
x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²
x = 12.0 m
"60 kg" is not a weight. It's a mass, and it's always the same
no matter where the object goes.
The weight of the object is
(mass) x (gravity in the place where the object is) .
On the surface of the Earth,
Weight = (60 kg) x (9.8 m/s²)
= 588 Newtons.
Now, the force of gravity varies as the inverse of the square of the distance from the center of the Earth.
On the surface, the distance from the center of the Earth is 1R.
So if you move out to 5R from the center, the gravity out there is
(1R/5R)² = (1/5)² = 1/25 = 0.04 of its value on the surface.
The object's weight would also be 0.04 of its weight on the surface.
(0.04) x (588 Newtons) = 23.52 Newtons.
Again, the object's mass is still 60 kg out there.
___________________________________________
If you have a textbook, or handout material, or a lesson DVD,
or a teacher, or an on-line unit, that says the object "weighs"
60 kilograms, then you should be raising a holy stink.
You are being planted with sloppy, inaccurate, misleading
information, and it's going to be YOUR problem to UN-learn it later.
They owe you better material.
Answer:
you will get huge electricity bills ............
