Answer:
a) 145.6kgm^2
b) 158.4kg-m^2/s
c) 0.76rads/s
Explanation:
Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation
(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and
(c) the angular speed of the merry-go-round and child after the child has jumped on.
a) From I = MK^2
I = (160Kg)(0.91m)^2
I = 145.6kgm^2
b) The magnitude of the angular momentum is given by:
L= r × p The raduis and momentum are perpendicular.
L = r × mc
L = (1.20m)(44.0kg)(3.0m/s)
L = 158.4kg-m^2/s
c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:
L = Iw
158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]
w = 158.6/208.96
w = 0.76rad/s
Answer:
The other angle is 120°.
Explanation:
Given that,
Angle = 60
Speed = 5.0
We need to calculate the range
Using formula of range
...(I)
The range for the other angle is
....(II)
Here, distance and speed are same
On comparing both range
Hence, The other angle is 120°
Answer:
The final pressure of the gas is 9.94 atm.
Explanation:
Given that,
Weight of argon = 0.16 mol
Initial volume = 70 cm³
Angle = 30°C
Final volume = 400 cm³
We need to calculate the initial pressure of gas
Using equation of ideal gas
Where, P = pressure
R = gas constant
T = temperature
Put the value in the equation
We need to calculate the final temperature
Using relation pressure and volume
Hence, The final pressure of the gas is 9.94 atm.
Answer:
Added sugars, Saturated fat, Trans fat.
Explanation:
Answer:
V = a * t = 9.8 m/s^2 * 2.3 s = 22.5 m/s velocity after 2.3 s
S = 1/2 g t^2 since initial speed is zero
S = 1/2 * 9.8 m/s^2 * 5.29 s^2 = 25.9 m
