The formula is Ke = 1/2 m v^2
The two of them together have a Ke of mv^2. So you either increase m or v. That's what makes the problem difficult. He can do D or B. We have to choose.
A is no solution. The Ke goes down because Paul loses Ivan's mass.
C is out of the question 3 meters/sec is a big reduction from 5 m/s. So now what do we do about B and D?
The question is what does the third person add. The tandoms I've peddled only allow for 1 or 2 people to add to the motion. So the third person only adds mass. He does not have a v that he is contributing to. To say that he is going 5m/s is true, but he's not contributing anything to that motion.
I pick B, but it is one of those questions that the correctness of it is in the head of the proposer. Be prepared to get it wrong. Argue the point politely if you agree with me, but back off as soon as you have presented your case.
B <<<<====== answer.
Answer:
F = 39.2 N (hand force) and N = 68.6 N (shoulder force)
Explanation:
In this exercise we must use the rotational and translational equilibrium conditions, we have several forces: the weight (W) of the pole applied at its geometric center, the load (w1) at one end, the shoulder support (N) 60 cm from the load and hand force (F) at the other end of the pole
Let's set the reference system at the fit point of the shoulder
∑ τ = 0
We will assume that the counterclockwise turns are positive
w₁ 0.60 + W 0.1 + F₁ 0 - F 0.4 = 0
all distances are measured from the support of the man (x₀ = 0.60 m)
F = (w₁ 0.60 + W 0.1) / 0.4
F = (m₁ 0.6 + m 0.1) g / 0.4
let's calculate
F = (2.6 0.6 + 0.4 0.1) 9.8 / 0.4
F = 39.2 N
this is the force that the hand must exert to keep the system in balance
We apply the translational equilibrium condition
-w₁ -W + N - F = 0
N = w₁ + W + F
N = (m₁ + m) g + F
let's calculate
N = (2.6 + 0.4) 9.8 + 39.2
N = 68.6 N
Impulse equals Change in Momentum
F = average applied force = to be determined
Δt = time during which the force is applied = 0.50 s
m = mass = 1,700 kg
Δp = change in momentum = to be determined
Δv = change in velocity = to be determined
v1 = initial velocity = 50.0 km/h = 50,000 m/h = 13.9 m/s
v2 = final velocity = 0.00 km/h = 0.00 m/s
F∙Δt = Δp
F∙Δt = m∙Δv
F∙Δt = m∙(v2 - v1)
F = m∙(v2 - v1) / Δt
F = 1,700 kg∙(0.00 m/s - 13.9 m/s) / 0.50 s
<span>F = -47,222 N The negative sign means that the force vector is </span>
<span>applied AGAINST the momentum vector of the rhinoceros.</span>
DE which is the differential equation represents the LRC series circuit where
L d²q/dt² + Rdq/dt +I/Cq = E(t) = 150V.
Initial condition is q(t) = 0 and i(0) =0.
To find the charge q(t) by using Laplace transformation by
Substituting known values for DE
L×d²q/dt² +20 ×dq/dt + 1/0.005× q = 150
d²q/dt² +20dq/dt + 200q =150
Answer:
Two marbles are launched at t = 0 in the experiment illustrated in the figure below. Marble 1 is launched horizontally with a speed of 4.20 m/s from a height h = 0.950 m. Marble 2 is launched from ground level with a speed of 5.94 m/s at an angle above the horizontal. (a) Where would the marbles collide in the absence of gravity? Give the x and y coordinates of the collision point. (b) Where do the marbles collide given that gravity produces a downward acceleration of g = 9.81 m/s2? Give the x and y coordinates.
Explanation:
i want the answer i don't know
