A student measured the density of Galena to be 7.9g/cm3 however the known density of Galena is 7.6g/cm3 . Calculate the percent
2 answers:
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Answer:
-3413 ft/s2
Explanation:
We need to know the velocity with which he landed on the snow.
He 'dropped' from 512 feet. This is the displacement. His initial velocity is 0 and the acceleration of gravity is 32 ft/s2.
We use the equation of mition
v and u are the initial and final velocities, a is the acceleration and s is the displacement. Putting the appropriate values
This is the final velocity of the fall and becomes the initial velocity as he goes into the snow.
In this second motion, his final velocity is 0 because he stops after a displacement of 4.8 ft. We use the same equation of motion but with different values. This time, , v = 0 and s = 4.8 ft.
Note that this is negative because it was a deceleration, that is, his velocity was decreasing.
Answer:
h = 2 R (1 +μ)
Explanation:
This exercise must be solved in parts, first let us know how fast you must reach the curl to stay in the
let's use the mechanical energy conservation agreement
starting point. Lower, just at the curl
Em₀ = K = ½ m v₁²
final point. Highest point of the curl
= U = m g y
Find the height y = 2R
Em₀ = Em_{f}
½ m v₁² = m g 2R
v₁ = √ 4 gR
Any speed greater than this the body remains in the loop.
In the second part we look for the speed that must have when arriving at the part with friction, we use Newton's second law
X axis
-fr = m a (1)
Y Axis
N - W = 0
N = mg
the friction force has the formula
fr = μ N
fr = μ m g
we substitute 1
- μ mg = m a
a = - μ g
having the acceleration, we can use the kinematic relations
v² = v₀² - 2 a x
v₀² = v² + 2 a x
the length of this zone is x = 2R
let's calculate
v₀ = √ (4 gR + 2 μ g 2R)
v₀ = √4gR( 1 + μ)
this is the speed so you must reach the area with fricticon
finally have the third part we use energy conservation
starting point. Highest on the ramp without rubbing
Em₀ = U = m g h
final point. Just before reaching the area with rubbing
= K = ½ m v₀²
Em₀ = Em_{f}
mgh = ½ m 4gR(1 + μ)
h = ½ 4R (1+ μ)
h = 2 R (1 +μ)