Answer:
The coefficient of static friction is 0.29
Explanation:
Given that,
Radius of the merry-go-round, r = 4.4 m
The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.
We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.
v is the speed of cat, 
So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.
Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
- momentum
- Yes, if the elephant is standing still.
- Fullback
- impulse acting on it.
- 2.25 N∙s
- A cannon firing.
- Inelastic
- it stays the same
- When the cue ball contacts the other balls, momentum is transferred causing them to gain momentum and speed.
- less than 3 m/s
<h3><u><em>
these are all correct i got an 100%</em></u><em><u> </u></em></h3>
Answer:
B = 62.9 N
Explanation:
This is an exercise on Archimedes' principle, where the thrust force equals the weight of the liquid
B = ρ g V
write the equilibrium equation
T + B -W = 0
B = W- T (1)
use the density to write the weight
ρ = m / V
m = ρ V
W = ρ g V
substitute in 1
B = m g -T
B = 
g V - T
To finish the calculation, the density of the material must be known, suppose it is steel \rho_{body} = 7850 kg / m³
calculate
B = 7850 9.8 1.20 10⁻³ - 29.4
B = 92.3 - 29.4
B = 62.9 N
Answer:
221.17 kJ
Explanation: Note the heat of vaporization is in kJ/mol,then to determine the number of moles of water: divide the mass by 18. Then multiply the number of moles by the molar heat of vaporization of water.
N = 97.6 ÷ 18
Q=molar heat *moles
Q = (40.79) * (97.6 ÷ 18)
This is approximately 221.17 kJ
