Answer:
519.62 m/s
Explanation:
Applying,
v = √(T/m').............. Equation 1
Where v = velocity of the wave, T = Tension on the string, m' = mass per unit length of the string
From the question,
Given: T = 1350 N, m' = 0.005 kg/m
Substitute these values into equation 1
v = √(1350/0.005)
v = √(270000)
v = 519.62 m/s
Answer:
Explanation:
Using the formula to calculate the maximum height
H = u²/2g
u is the initial velocity = 18m/s
g is the acceleration due to gravity = 9.81
H = 18²/2(9.81)
H = 324/19.62
H = 16.51m
The maximum height in which the ball reached from the ground = 85m + 16.51m = 101.5m
b) Time of fight = 2u/g
T = 2(18)/g
T = 36/9.81
T = 3.67s
It took the ball 3.67s later to reach the ground
3) To get the final velocity of the ball as it hits the ground, we need to calculate the horizontal component of the velocity,
Ux = Ucosθ
ux = 18cos 90 (angle of launch is 90 since the ball is thrown vertically upwards)
Ux = 18(0)
Ux = 0m/s
Hence the final velocity of the ball as it hits the ground is 0m/s
Answer:
Explanation:
As we know that flux of magnetic field from the closed loop is given by
now we will have
now we have
EMF = 10 Volts
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