Answer:
The spring constant = 9.25 N/m
Explanation:
The equation of an object attached to a spring that is oscillating is
T = 2π√(m/k)
Where T = period of the oscillation, m = mass of the object, k = spring constant.
Making k the subject of the equation,
k = 4π²m/T²......................... Equation 1
Note: Period(T) is the time taken to complete one oscillation
Given: T = t/10 = 9.0/10 = 0.9 s, m = 190 g = 0.19 kg.
Constant: π = 3.14
Substitute these values into equation 1.
k = 4(3.14)²(0.19)/0.9²
k = 7.4933/0.81
k = 9.25 N/m
Thus the spring constant = 9.25 N/m
Answer:
Thrust acts on the accelerated object in the direction opposite to the applied force hence it accelerates the object in the direction opposite to the applied force. ... Its magnitude is equal to that of applied force. It always increases the velocity of the object.
Explanation:
The electric output of the plant is 48.19 MW
First we need to calculate the water power, it is given by the formula
WP=ρQgh
Here, ρ=1000 kg/m3 is density of water,Q is the flow rate, g is the gravity, and h is the water head
Therefore, WP=1000*65*9.81*90=57388500 W=57.38 MW
Now the overall efficiency of the hydroelectric power plant is given as
η=
Plugging the values in the above equation
0.84=EP/57.38
EP=48.19 MW
Therefore, the electric output of the plant is 48.19 MW.
<span>An object roating at one revokution per second has an angular velocity of 360 degrees per second or 2pi radians per second. This is found by taking the number of revolutions over a period of time and than dividing by the chosen period of time to get the velocity. There are 360 degrees or 2pi radians in one revolution.</span>
