Answer:
Explanation:
The bead is moving on a vertical circular path so it must have a centripetal force towards the centre.
This force is equal to m v² / r
v is velocity of bead and r is radius of the circular path.
The vertical hoop is also rotating about a vertical axis passing through the centre at frequency f so the bead will experience a cetrifugal force due to rotation of the hoop. Its value is
m ω² r . Only at the point o degree and 180 degree , these forces are opposite to each other so at these points , the bead will be in equilibrium .
mv² / r = m ω² r
v² = ω² r²
v = ω r
= 2π f r
= 2 x 3.14 x 2 x 0.22
v = 2.76 m /s
For the bead to rise upto point θ = 90 degree , height achieved is radius R
required velocity = √ 2gR
= √ 2x 9.8x.22
= 2.076 m/s
This velocity is less than the velocity calculated earlier so the bead will be able to ride the required height.
v = 2.76 m/s
Answer:
The thrust is 
Explanation:
Given that,
Mass of gas, 
The rate at which the gas is expelling, 
We need to find the thrust produced by the gas.
We know that force is equal to the rate of change of momentum. So,

Also, p = mv

So,

So, the thrust is 
It contains no large maria
Answer:
the equilibrium wage rate is 10 and the equilibrium quantity of labor is 1000 workers
Explanation:
The equilibrium wage rate and the equilibrium quantity of labor are found as the point where the equation of demand intercepts the equation of supply, so the equilibrium quantity of labor is:

15 - (1/200) L = 5 + (1/200) L
15 - 5 = (1/200) L + (1/200) L
10 = (2/200) L
(10*200)/2 = L
1000 = L
Then, the equilibrium wage rate is calculated using either the equation of demand for labor or the equation of supply of labor. If we use the equation of demand for labor, we get:
W = 15 - (1/200) L
W = 15 - (1/200) 1000
W = 10
Finally, the equilibrium wage rate is 10 and the equilibrium quantity of labor is 1000 workers
Answer:
v_average = 15 m / s
Explanation:
The average speed can be found in two ways,
* taking the distance traveled and divide it by the time spent
* taking the velocities in each time interval and then finding the weighted average by the time fraction
v_average = 1 / t_total ∑
vi ti
Let's apply this last equation
Total time is
t = t₁ + t₂
t = 10 + 10 = 20 min
v_average = 10/20 10 + 10/20 20
v_average = 10/2 + 20/2
v_average = 15 m / s