At the lowest point on the Ferris wheel, there are two forces acting on the child: their weight of 430 N, and an upward centripetal/normal force with magnitude n; then the net force on the child is
∑ F = ma
n - 430 N = (430 N)/g • a
where m is the child's mass and a is their centripetal acceleration. The child has a linear speed of 3.5 m/s at any point along the path of the wheel whose radius is 17 m, so the centripetal acceleration is
a = (3.5 m/s)² / (17 m) ≈ 0.72 m/s²
and so
n = 430 N + (430 N)/g (0.72 m/s²) ≈ 460 N
Answer:
Hits per second=199 hit/s
Explanation:
#Given the angular velocity,
, radius of the record
and the distance between any two successive bumps on the groove as
.
The linear speed of the record in meters per second is:

#From
above, if the bumps are uniformly separated by 1m, then the rate at which they hit the stylus is:

Hence the bumps hit the stylus at around 199hit/s
Answer:
W = - 118.24 J (negative sign shows that work is done on piston)
Explanation:
First, we find the change in internal energy of the diatomic gas by using the following formula:

where,
ΔU = Change in internal energy of gas = ?
n = no. of moles of gas = 0.0884 mole
Cv = Molar Specific Heat at constant volume = 5R/2 (for diatomic gases)
Cv = 5(8.314 J/mol.K)/2 = 20.785 J/mol.K
ΔT = Rise in Temperature = 18.8 K
Therefore,

Now, we can apply First Law of Thermodynamics as follows:

where,
ΔQ = Heat flow = - 83.7 J (negative sign due to outflow)
W = Work done = ?
Therefore,

<u>W = - 118.24 J (negative sign shows that work is done on piston)</u>