Here We can use principle of angular momentum conservation
Here as we know boy + projected mass system has no external torque
Since there is no torque so we can say the angular momentum is conserved
now we know that
m = 2 kg
v = 2.5 m/s
L = 0.35 m
I = 4.5 kg-m^2
now plug in all values in above equation
so the final angular speed will be 0.37 rad/s
Answer:
121.3 cm^3
Explanation:
P1 = Po + 70 m water pressure (at a depth)
P2 = Po (at the surface)
T1 = 4°C = 273 + 4 = 277 K
V1 = 14 cm^3
T2 = 23 °C = 273 + 23 = 300 K
Let the volume of bubble at the surface of the lake is V2.
Density of water, d = 1000 kg/m^3
Po = atmospheric pressure = 10^5 N/m^2
P1 = 10^5 + 70 x 1000 x 10 = 8 x 10^5 N/m^2
Use the ideal gas equation
By substituting the values, we get
V2 = 121.3 cm^3
Thus, the volume of bubble at the surface of lake is 121.3 cm^3.
Answer:
No one is right
Explanation:
John Case:
The function is defined between -1 and 1, So it is not possible obtain a value
greater.
In addition, if you move the function cosine a T Value, and T is the Period, the function take the same value due to the cosine is a periodic function.
Larry case:
Is you have , the domain of this is [0,2].
it is equivalent to adding 1 to the domain of the , and its mean that the function
, in general, is not greater than
.