Answer:
15 N and 3.061
Explanation:
From the question,
The minimum force of friction to keep the book from sliding = 15 N.
using
F = mgμ................. Equation 1
Where F = Frictional Force, m = mass of the book, g = acceleration due to gravity, μ = coefficient of friction.
make μ the subject of the equation
μ = F/mg............... Equation 2
Given: F = 15 N, m = 0.5 kg, g = 9.8 m/s²
Substitute into equation 2
μ = 15(0.5×9.8)
μ = 15/4.9
μ = 3.061
Hence the coefficient of friction to keep the book from sliding = 3.061
For this case we first think that the skateboard and the child are one body.
We have then:
1 = jug
2 = skateboard + boy
By conservation of the linear amount of movement:
M1V1i + M2V2i = M1V1f + M2V2f
Initial rest:
v1i = v2i = 0
0 = M1V1f + M2V2f
Substituting values
0 = (7.8) (3.2) + (M2) (- 0.65)
0 = 24.96 + M2 (-0.65)
-24.96 = (-0.65) M2
M2 = (-24.96) / (- 0.65) = 38.4 kg
Then, the child's mass is:
M2 = Mskateboard + Mb
Clearing:
Mb = M2-Mskateboard
Mb = 38.4 - 1.9
Mb = 36.5 Kg
answer:
the boy's mass is 36.5 Kg
1) 2 kilograms converted into grams is 2000
2) 5200 meters converted into kilometers is 5.2
3) 20 centimeters converted into meters is 0.2
<span><u>Answer</u>
c). random internal motion of atoms and molecules.
<u>Explanation </u>
Diffusion is the movement of particles from a region of high concentration to region of low concentration. The rate of diffusion can be increased by increasing amount of temperature but this is not its primary cause. Diffusion is primarily caused by movement of molecules or atoms in a substances. Form the choices given, the correct answer is c. random internal motion of atoms and molecules.
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Answer:
a. 0.342 kg-m² b. 2.0728 kg-m²
Explanation:
a. Since the skater is assumed to be a cylinder, the moment of inertia of a cylinder is I = 1/2MR² where M = mass of cylinder and r = radius of cylinder. Now, here, M = 56.5 kg and r = 0.11 m
I = 1/2MR²
= 1/2 × 56.5 kg × (0.11 m)²
= 0.342 kgm²
So the moment of inertia of the skater is
b. Let the moment of inertia of each arm be I'. So the moment of inertia of each arm relative to the axis through the center of mass is (since they are long rods)
I' = 1/12ml² + mh² where m = mass of arm = 0.05M, l = length of arm = 0.875 m and h = distance of center of mass of the arm from the center of mass of the cylindrical body = R/2 + l/2 = (R + l)/2 = (0.11 m + 0.875 m)/2 = 0.985 m/2 = 0.4925 m
I' = 1/12 × 0.05 × 56.5 kg × (0.875 m)² + 0.05 × 56.5 kg × (0.4925 m)²
= 0.1802 kg-m² + 0.6852 kg-m²
= 0.8654 kg-m²
The total moment of inertia from both arms is thus I'' = 2I' = 1.7308 kg-m².
So, the moment of inertia of the skater with the arms extended is thus I₀ = I + I'' = 0.342 kg-m² + 1.7308 kg-m² = 2.0728 kg-m²
