Answer:
m = 105.37 kg
Explanation:
We are given;
Mass of man; m = 113 kg
Length of boat = 6.3m
Now, The position of the center of mass will not change during the motion of the man.
Thus,
X_g,i = X_g,f
So,
[113(6.3) + ma]/(113 + m) = [113(3.26) + m(a +3.26)]/(113 + m)
113 + m will cancel on both sides to give;
113(6.3) + ma = [113(3.26) + m(a +3.26)]
711.9 + ma = 368.38 + ma + 3.26m
ma will cancel out to give;
711.9 - 368.38 = 3.26m
343.52/3.26 = m
m = 105.37 kg
Answer:
Explanation:
a = 4ms⁻², Vf = 180 m/s & Vi = 140m/s
a =
4 = 
t = 40/4
t = 10sec
To Measure Distance Use third Equation of Motion:
2aS = Vf²-Vi²
S = 
S = 12800/8 = 1600m
Answer:
R= 602 .11 N
Explanation:
The horizontal component of tension T will give reaction of the wall and the vertical component of T will balance the weight of of the climber .
T cos32 = R
710 x .848 = R
R= 602 .11 N .
Answer:
a.) τ = 2.85 s b.) Q = 3.19 * 10^-5 C c.) t = 1.691 s
Explanation:
So we are told that it is a RC circuit. We are told
= 12.0 V, R = 1.07 MΩ and C = 2.66 µF.
a.) The time constant for RC circuit, τ = RC. Substituting our known values we get:
τ = RC where R = (1.07 * 10 ^ 6)Ω and C = (2.66 * 10 ^ -6) F
τ = (1.07 * 10 ^ 6)Ω * (2.66 * 10 ^ -6) F = 2.8462 s ≈ 2.85 s
τ = 2.85 s
b.) The relationship between capacitance, potential, charge is given:
![Q = CV[1-e^{-t/RC} ]](https://tex.z-dn.net/?f=Q%20%3D%20CV%5B1-e%5E%7B-t%2FRC%7D%20%5D)
The capacitor is fully charge when t approaches infinity, therefore:
![Q = \lim_{t \to \infty} a_n CV[1-e^{-t/RC} ]](https://tex.z-dn.net/?f=Q%20%3D%20%20%5Clim_%7Bt%20%5Cto%20%5Cinfty%7D%20a_n%20CV%5B1-e%5E%7B-t%2FRC%7D%20%5D)
When t approaches infinity, the term e becomes very small (e^-∞ = 0), therefore we can simplify the equation and plug in our values
![Q = (2.66*10^{-6}) F * (12.0)V *[1 - 0] = 3.192 * 10^{-5}](https://tex.z-dn.net/?f=Q%20%3D%20%282.66%2A10%5E%7B-6%7D%29%20F%20%2A%20%2812.0%29V%20%2A%5B1%20-%200%5D%20%3D%203.192%20%2A%2010%5E%7B-5%7D)
Q = 3.19 * 10^-5 C
c.) Using the same equation as before, we can substitute Q in and solve for Q:
![(14.3 * 10 ^ 6) C = (2.66*10^{-6})F *(12.0)V*[1-e^{-t/(2.85s)}]\\0.552 = e^{-t/(2.85s)}\\t = -1 * 2.85 * ln(0.552) \\t = 1.69120678 s](https://tex.z-dn.net/?f=%2814.3%20%2A%2010%20%5E%206%29%20C%20%3D%20%282.66%2A10%5E%7B-6%7D%29F%20%2A%2812.0%29V%2A%5B1-e%5E%7B-t%2F%282.85s%29%7D%5D%5C%5C0.552%20%3D%20e%5E%7B-t%2F%282.85s%29%7D%5C%5Ct%20%3D%20-1%20%2A%202.85%20%2A%20ln%280.552%29%20%5C%5Ct%20%3D%201.69120678%20s)
t = 1.691 s
Hope this helps! I'm not sure what the units you want, so convert to the desired units.