Answer:
since -6 lasted for 5 seconds, multiplying both would result in -30
3 lasted for 10 seconds, so multiplying both would give +30
average = ( 30 + (-30) ) / 2
30 -30 is already equal to zero, so the answer should be 0
Answer: µ=0.205
Explanation:
The horizontal forces acting on the ladder are the friction(f) at the floor and the normal force (Fw) at the wall. For horizontal equilibrium,
f=Fw
The sum of the moments about the base of the ladder Is 0
ΣM = 0 = Fw*L*sin74.3º - (25.8kg*(L/2) + 67.08kg*0.82L)*cos74.3º*9.8m/s²
Note that it doesn't matter WHAT the length of the ladder is -- it cancels.
Solve this for Fw.
0= 0.9637FwL - (67.91L)2.652
Fw=180.1/0.9637
Fw=186.87N
f=186.81N
Since Fw=f
We know Fw, so we know f.
But f = µ*Fn
where Fn is the normal force at the floor --
Fn = (25.8 + 67.08)kg * 9.8m/s² =
910.22N
so
µ = f / Fn
186.81/910.22
µ= 0.205
- m1=1500kg
- m_2=3000kg
- v_1=5m/s
- v_2=7m/s
Using law of conservation of momentum
Given Information:
KEa = 9520 eV
KEb = 7060 eV
Electric potential = Va = -55 V
Electric potential = Vb = +27 V
Required Information:
Charge of the particle = q = ?
Answer:
Charge of the particle = +4.8x10⁻¹⁸ C
Explanation:
From the law of conservation of energy, we have
ΔKE = -qΔV
KEb - KEa = -q(Vb - Va)
-q = KEb - KEa/Vb - Va
-q = 7060 - 9520/27 - (-55)
-q = 7060 - 9520/27 + 55
-q = -2460/82
minus sign cancels out
q = 2460/82
Convert eV into Joules by multiplying it with 1.60x10⁻¹⁹
q = 2460(1.60x10⁻¹⁹)/82
q = +4.8x10⁻¹⁸ C
