Given that,
Mass of trackler, m₁ = 100 kg
Speed of trackler, u₁ = 2.6 m/s
Mass of halfback, m₂ = 92 kg
Speed of halfback, u₂ = -5 m/s (direction is opposite)
To find,
Mutual speed immediately after the collision.
Solution,
The momentum of the system remains conserved in this case. Let v is the mutual speed after the collision. Using conservation of momentum as :
So, the mutual speed immediately after the collision is 1.04 m/s but in opposite direction.
Answer:
kg m/s
Explanation:
e = Charge = C
V = Voltage = 
c = Speed of light = m/s
Momentum is given by
The unit of MeV/c in SI fundamental units is kg m/s
Answer:
Por ela ter batido na trave, não tem como voltar 2x mais forte, por que toda ação correspondente a uma reação de igual intensidade, mas que atua no sentido oposto
Explanation:
A and B are equivalent. That's one way instruments are often grouped. (the "sopranos", the "altos", the "bass")
C is another way instruments are often grouped; (the "woods", the "brass")
D is another way instruments are often grouped; (the "strings", the "percussions")
I'm actually going ahead in the book (DC Circuits) so this isn't really homework but I figured the tag was appropriate....the name of the chapter is Ohm's Law and Watt's Law.
<span>Problem: Calculate the power dissipated in the load resistor, R, for each of the circuits.Circuit (a): V = 10V; I = 100mA; R = ?; Since I know
V and
I use formula
P = IV: P = IV = (100mA)(10V) = 1 W.</span>
The next question is what I'm not sure about:
Question: What is the power in the circuit (a) above if the voltage is doubled? (Hint: Consider the effect on current).
What I did initially was: P = IV = (100mA)(2V) = 2 W
But then I looked at the answer and it said 4 W, then I looked at the Hint again. Then I remembered in the book early on it said "If the voltage increases across a resistor, current will increase."
So question is: When solving problems I have to increase (or decrease) current (I) every time voltage (V) is increased (decreased) in a problem, right? How about the other way around, when increasing current (I), you need to increase voltage (V). I'm pretty sure that's how they got 4 W, but want to make sure before I head to the next section of the book.
P = IV = (200mA)(2V) = 4 W
