Answer:
TEJ as this is a thing you wont get
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
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<span><span>The
best and most correct answer among the choices provided by the question is </span>B.-2.71 V.</span>
Mg2+(aq) + 2e- -> Mg(s) E=-2.37 V
Cu2+(aq) + 2e- -> Cu(s) E =+ 0.34 V
Since Cu is acting as the anode, the equation needs to be
reversed.
Cu(s) -> Cu2+(aq) + 2e- E =- 0.34 V
Ecell= -2.37 V+ (- 0.34 V) = -2.71 V
<span><span>
</span><span>Hope my answer would be a great help for you. </span> </span>
<span> </span>
Answer:
Explanation:
Given
distance traveled by Passenger train is 
distance traveled by Freight train is 
Both take same time to travel
suppose speed of freight train is x mph
so speed of passenger train is 20+x mph
we know time




speed of passenger train
Answer:
I think so but i could be wrong..
Explanation: