W=MG
w is weight
m is mass
g is gravity
W=(100 kg)(9.8 m/s)
W= 980 N
hope this helps
Let us evaluate the given assumptions according to the kinetic theory for an ideal gas.
a.
The motion of one particle is unaffected by other particles unless the particles collide.
TRUE. The particles are in random motion unless they collide.
b.
The forces of attraction among particles keep the particles close together.
FALSE. No forces act between particles except during collision.
c.
Under ordinary conditions, forces of attraction between particles can be ignored.
TRUE.
Answer: Statement b is false because it is not an assumption.
The solution for this problem:
Given:
f1 = 0.89 Hz
f2 = 0.63 Hz
Δm = m2 - m1 = 0.603 kg
The frequency of mass-spring oscillation is:
f = (1/2π)√(k/m)
k = m(2πf)²
Then we know that k is constant for both trials, we have:
k = k
m1(2πf1)² = m2(2πf2)²
m1 = m2(f2/f1)²
m1 = (m1+Δm)(f2/f1)²
m1 = Δm/((f1/f2)²-1)
m 1 = 0.603/
(0.89/0.63)^2 – 1
= 0.609 kg or 0.61kg or 610 g
Answer:
a) i = -9.63 cm
, h ’= .0.24075 cm erect
b) i = 259.74 cm
,
Explanation:
For this exercise let's start by finding the focal length of the lens
1 / f = (n-1) (1 / R₁ - 1 / R₂)
1 / f = (1.70 -1)) 1 / ∞ - 1/13)
1 / f = 0.0538
f = - 18.57 cm
Now we can use the constructor equation
1 / f = 1 / o + 1 / i
1 / i = 1 / f - 1 / o
1 / i = -1 / 18.57 -1/20
1 / i = -0.1038 cm
I = -9.63 cm
For the height of the
image let's use magnification
m = h '/ h = - i / o
h ’= -h i / o
h ’= - 0.5 (-9.63) / 20
h ’= .0.24075 cm
b) we invert the lens
The focal length is
1 / f = (1.70 -1) (1/13 - 1 / int)
1 / f = 0.0538
f = 18.57 cm
1 / i = 1 / f -1 / o
1 / I = 1 / 18.57 - 1/20
1 / I = 3.85 10-3
i = 259.74 cm
h ’= - 0.5 259.74 / 20
h ’= 6.4935 cm
Answer:
The lever is a movable bar that pivots on a fulcrum attached to a fixed point. The lever operates by applying forces at different distances from the fulcrum, or a pivot. As the lever rotates around the fulcrum, points farther from this pivot move faster than points closer to the pivot.
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