Answer:
Part (i)
Z = 39.06 ohm
Part (ii)
R = 21.7 ohm
Explanation:
a) here we know that
maximum value of EMF = 125 V
maximum value of current = 3.20 A
now by ohm's law we can find the impedence as
now we will have
Part b)
Now we also know that
now we have
Missing figure and missing details can be found here:
<span>http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....
</span>
Solution:
(a) The work done by the spring is given by
where k is the elastic constant of the spring and
is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have
(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
where the negative sign is given by the fact that
points in the opposite direction of the displacement of the cart, and where
therefore, the work done by the weight is
Answer
given,
mass of ball, m = 57.5 g = 0.0575 kg
velocity of ball northward,v = 26.7 m/s
mass of racket, M = 331 g = 0.331 Kg
velocity of the ball after collision,v' = 29.5 m/s
a) momentum of ball before collision
P₁ = m v
P₁ = 0.0575 x 26.7
P₁ = 1.535 kg.m/s
b) momentum of ball after collision
P₂ = m v'
P₂ = 0.0575 x (-29.5)
P₂ = -1.696 kg.m/s
c) change in momentum
Δ P = P₂ - P₁
Δ P = -1.696 -1.535
Δ P = -3.231 kg.m/s
d) using conservation of momentum
initial speed of racket = 0 m/s
M u + m v = Mu' + m v
M x 0 + 0.0575 x 26.7 = 0.331 x u' + 0.0575 x (-29.5)
0.331 u' = 3.232
u' = 9.76 m/s
change in velocity of the racket is equal to 9.76 m/s
Answer:
L/2
Explanation:
Neglect any air or other resistant, for the ball can wrap its string around the bar, it must rotate a full circle around the bar. This means the ball should be able to swing to the top position where it's directly above the bar. By the law of energy conservation, this happens when the ball is at the same level as where it's previously released vertically. It means the swinging radius around the bar must be at least half of the string length.
So the distance d between the bar and the pivot should be at least L/2
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