The appearance of the organism and its structures, because the fossils are basically worn out parts of the organism. IT IS basically the organism.
For the 5.0Kg block with the force of 15 Newton, the coefficient of static friction (μ) comes out to be 0.306. Applying the formula F = μmg.
Static friction acts on stationary body/body at rest. Let μ be the coefficient of static friction between the block and the horizontal floor.
Using the formula: F = μmg
Where,
F = Force , m= mass of the block and g = gravity.
and values of: m= 5.0Kg, F= 15.0N, g= 9.8m/s²
We can get: μ = F/(mg)
μ = 15/49
μ =0.306
Therefore, coefficient of static friction (μ) = 0.306.
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Answer:
infrared part of the spectrum
Explanation:
brown dwarfs are relatively cool and have temperatures of about 2000 K emitting their light in the infrared and practically none in the ultraviolet part of the spectrum.
True i believe. It says my comment has to be longer so i’m just typingggg
Answer:
Explanation:
First of all we shall calculate the velocity of composite mass . Let it be v . Applying law of conservation of momentum
mu - MU = ( m + M ) v
v = mu - MU / ( m + M )
loss of kinetic energy
= 1/ 2 mu² + 1/2 MU² - 1/2 ( M +m ) v²
= 1/ 2 mu² + 1/2 MU² - 1/2 ( M +m ) (mu - MU)² / ( m + M )²
= 1/ 2 mu² + 1/2 MU² - 1/2 (mu - MU)² / ( m + M )
= 1/2 [ m²u² + mMu² +mMU² + m²U² - m²u² - M²U² - 2 muMU ] / ( m + M )
= 1 / 2 [ mMu² + mMU² - 2 muMU ] / ( m + M )
= 1 / 2mM [ (u² + U² - 2 uU) / ( m + M )]
= 1/2 mM x k
where
k = [ (u² + U² - 2 uU) / ( m + M )]
Given
m = .004 kg
M = 4 kg
u = 890 ms⁻¹
U = 7 ms⁻¹
k = ( 890² + 7² - 2 x 890 x 7 ) / 4.004
= ( 792100 + 49 - 12460 ) / 4.004
= 194727.52
loss of kinetic energy
= 1/2 mM x k
= .5 x .004 x 4 x 194727.52
= 1557.82 J .
