Answer:
proteins carbohydrates and fats
Explanation:
OF COURSE !
The gravitational force between two objects doesn't ONLY depend on the product of their masses. It also depends on the distance between them.
I'm not even going to work out the numbers for my example. I'm just going to state without proof that at the top of the 2nd frame, the gravitational force between you and your bowling ball is greater than the gravitational force between you and the whole darn Andromeda galaxy ! My reasoning is based on the fact that your bowling ball is maybe 1 foot from your center of mass, whereas the Andromeda galaxy is more like 2.5 million light years from it. That right there is going to give your bowling ball a big advantage when it comes to gravity !
Answer:
Explanation:
First of all, we analyze the system of blocks before starting to move.
Hence, the incline angle θ for which both blocks begin to slide is 10.20°.
Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.

Where:



Therefore, the required stretch or compression in the connecting spring is 0.10 ft.
I hope it helps you!
Answer:
(a) 1462.38 m/s
(b) 2068.13 m/s
Explanation:
(a)
The Kinetic energy of the atom can be given as:
K.E = (3/2)KT
where,
K = Boltzman's Constant = 1.38 x 10⁻²³ J/k
K.E = Kinetic Energy of atoms = 343 K
T = absolute temperature of atoms
The K.E is also given as:
K.E = (1/2)mv²
Comparing both equations:
(1/2)mv² = (3/2)KT
v² = 3KT/m
v = √[3KT/m]
where,
m = mass of Helium = (4 A.M.U)(1.66 X 10⁻²⁷ kg/ A.M.U) = 6.64 x 10⁻²⁷ kg
v = RMS Speed of Helium Atoms = ?
Therefore,
v = √[(3)(1.38 x 10⁻²³ J/K)(343 K)/(6.64 x 10⁻²⁷ kg)]
<u>v = 1462.38 m/s</u>
(b)
For double temperature:
T = 2 x 343 K = 686 K
all other data remains same:
v = √[(3)(1.38 x 10⁻²³ J/K)(686 K)/(6.64 x 10⁻²⁷ kg)]
<u>v = 2068.13 m/s</u>
When any boat displaces a weight of water equal to its own weight, it floats. This is called the "principle of flotation": A floating object displaces a weight of fluid equal to its own weight.