Answer:
When an object is in free fall, gravity increases its velocity by 9.8 m/s with every passing second.
Explanation:
Hope Helps :)
Answer:u can go to the moon and bottom of the ocean
Explanation:
Because the suits help them do both of that
Answer:
1.33 m
Explanation:
Tom weighs 600 N and is 2 meters from the pivot. So his moment is:
τ = Fr
τ = (600 N) (2 m)
τ = 1200 Nm
To balance the see-saw, Jeremy must create an equal and opposite moment.
τ = Fr
1200 Nm = (900 N) r
r = 1.33 m
So Jeremy must sit 1.33 m from the pivot on the other side.
The correct answer is Equal to the weight of the block.
We are aware that when an object floats, the water it displaces weighs the same as the weight of the floating body. The upthrust is now equivalent to the body weight in this situation. The beaker is overflowing with water in this instance. The body was then created to float into it. When it reaches equilibrium in water, it expels water that must have been the same weight as it. Over the edge, the liquid spilled, collecting on the tray.
Therefore, the amount of water gathered on the tray must match the weight of the person floating.
Learn more about weight here-
brainly.com/question/10069252
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Answer:
θ = 3.19 arc second
Explanation:
The resolution of a telescope is given by the rayleigh criterion, which establishes that two objects are separated if the principal maximum of diffraction of one of them coincides with the first minimum of diffraction of the second object, based on this the solution is given by the first diffraction minimum, the a slit is
a sin θ = m λ
with m = 1
in the case of circular apertures the equation must be found in polar coordinates, therefore a numerical constant is introduced
a sin θ = 1.22 λ
Angles are measured in radians and in these experiments they are small
sin θ = θ
θ= 1.22 λ / a
in this case a = 6.09 in, the wavelength is wrong = 550 10⁻⁹ m which is the maximum resolution of the human eye
l
et's reduce the magnitudes to the SI system
d = 6.09‘ 2.54 10⁻-2 m / 1 inch = 15.4686 10-2 m
let's calculate
θ = 1.22 550 10-9 / 15.468 10-2
θ = 15.5 10⁻⁶ rad
rad = 2.06 105 s
θ = 15.5 10⁻⁶ rad 2.06 105s/ 1 rad
θ = 3.19 s
