If you increase the voltage what will the current do
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The resulting change in momentum of the system will be +18.6 Ns. The momentum is conserved.
According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.
The given data in the problem is;
m is the mass =6.0 kg
t is the time interval=2 second
From Newton's second law;
From the graph;
The change in the momentum is;
Hence, the resulting change in momentum of the system will be +18.6 Ns.
To learn more about the law of conservation of momentum, refer;
#SPJ1
Answer:
time=4s
Explanation:
we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion
, where
max=
Given that, at i(2)=
⇒
⇒
⇒
Applying logarithm on both sides,
⇒
⇒
⇒
Now substitute
⇒
⇒
⇒
Applying logarithm on both sides,
⇒
⇒
⇒
now subs.
⇒
also
⇒
⇒
<u>Given data:</u>
m= 50 Kg,
W= m×g = 50 × 9.81 = 490.5 N
ramp angle (α) = 30 degrees,
coefficient of friction (μs) = 0.5773,
Push at an angle (Θ) = 40 degrees,
Determine: Push to get box move up (P)=?
From the figure,
Resolving the forces along the plane
W sinα + μs.R = P cos Θ --------------------- (i)
Resolving the forces perpendicular to the inclined plane
W cosα = R+Psin Θ => R= W cosα - Psin Θ -------------- (ii)
Solving (i) and (ii) and keeping <em>μs = tan Φ, Φ = Θ </em>
<em>Pmin = W sin( α +Θ )</em>
<em> = W[ sin α.Cos Θ + cos α.sin Θ]</em>
<em> = 490.5 [ (sin 30.cos40) + (cos30.sin 40)]</em>
<em> = 460.9 N</em>
<em>Minimum push required to move the box up the ramp is 460.9 N</em>
