Answer:
You are pulled towards that building. At the same time, that building is pulled towards you. Neither object creates enough gravitational force to really do anything. That is why you never notice any affect by either body, (you and a building).
Explanation:
You will surely get attracted towards the building.But it takes a lot of time depending on their masses.
This happens only when you are away from earth with that building.
Both of you will get attracted to it
if a third party with mass more than you or building is with you.
If it is on the earth.. Then the gravity between you and the building is negligible compared to the earth.Hence you will not get attracted towards the building in this case.
Answer:
Democritus
Explanation:
Democritus was a Greek philosopher who was the first person to use the term atom . However ,John Dalton was the first to adapt Democritus’ theory into the first modern atomic
model
Answer:
A = 2 cm
, λ = 8 cm
Explanation:
The amplitude of a wave is the maximum height it has, in this case the height is measured by the vertical ruler,
We are told the balance point is in the reading of 5 cm, that the maximum reading is 3 cm and the Minimum reading is 7 cm. Therefore, the distance from the ends of the ridge to the point of equilibrium is
d = 7-5 = 2 cm
d = 5-3 = 2 cm
A = 2 cm
The wavelength is the minimum horizontal distance for which the wave is repeated, that is measured by the horizontal ruler.
The initial reading for 4 cm and the final reading for 8 cm, this distance corresponds to a crest of the wave, the complete wave is formed by two crests whereby the wavelength is twice this value
Δx = 8-4 = 4 cm
λ = 2 Δx
λ = 8 cm
Answer:
Part a)

Part b)

Explanation:
Part A)
As we know that time period of the motion is given as

so we have



now at the point of maximum amplitude the force equation when Normal force is about to zero is given as

so we have



Part b)
Now if the amplitude of the SHM is 6.23 cm
and now at this amplitude if object will lose the contact then in that case again we have



so now we have


Answer:

Explanation:
As we know that we board in the car of ferris wheel at the bottom position
So we will have
final height of the car at angular displacement given as


here we know that


so we have

