Answer:
Hey!
Your answer should be D!
Explanation:
In a transformer Np / Ns is called the voltage ratio. If Ns is less than Np then Vs is less than Vp. This is called a step-down transformer as the voltage is reduced.
(source from google.com!)
The period of oscillation is T = 2 * pi * sqrt ( ( m2/3 + m1) / k )
<h3>What is period of oscillation?</h3>
This is the time in seconds it takes to complete one oscillation. where an oscillation is a repetitive to and fro motion. period if the inverse of frequency and both are basic when calculation motion in simple harmonic motion.
The period of oscillation is given as T
T = 2 * pi * sqrt ( m / k )
where
m = mass on this case mass of the spring will be inclusive to the mass of the block such that we have:
m1 = mass of the block
m2 = mass pf the spring
k = force constant of the spring
including the two masses to the period gives
T = 2 * pi * sqrt ( ( m2/3 + m1) / k )
Read more on period of oscillation here: brainly.com/question/22499336
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Answer:
The magnitude of gravitational force between two masses is .
Explanation:
Given that,
Mass of first lead ball,
Mass of the other lead ball,
The center of a large ball is separated by 0.057 m from the center of a small ball, r = 0.057 m
We need to find the magnitude of the gravitational force between the masses. It is given by the formula of the gravitational force. It is given by :
So, the magnitude of gravitational force between two masses is . Hence, this is the required solution.
Answer:
Vi = 8.28 m/s
Explanation:
This problem is related to the projectile motion.
As we know there are two components of motion associated with this, the horizontal component and vertical component.
The horizontal distance covered by the ball is
Vx*t = x
Vx*t = 5.3
Vx = 5.3/t eq. 1
Also we know that
Vx = Vicos(60)
Vx = Vi*0.5 eq. 2
equate eq. 1 and eq. 2
5.3/t = Vi*0.5
5.3/0.5 = Vi*t
Vi*t = 10.6 eq. 3
The vertical distance is
Vy = y1 + Vyi*t - 0.5gt²
also we know that
Vyi = Visin(60)
Vyi = Vi*0.866
It is given that V1 = 1.9 m and and Vy = 3 m is the vertical distance
3 = 1.9 + Vi*0.866*t - 0.5gt²
3 = 1.9 + Vi*0.866*t - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
1.1 = 0.866(Vi*t) - 4.9t²
0.866(Vi*t) = 4.9t² + 1.1
substitute Vi*t = 10.6 in above equation
0.866(10.6) = 4.9t² + 1.1
9.18 = 4.9t² + 1.1
4.9t² = 8.08
t² = 8.08/4.9
t² = 1.648
t = 1.28 sec
Finally, initial speed can be found by substituting the value of t into eq. 3
Vi*t = 10.6
Vi = 10.6/t
Vi = 10.6/1.28
Vi = 8.28 m/s