Answer:
○ 0
Explanation:
Even if the athelete runs four laps of a 400 m track, its displacement will be 0 because the displacement is the shortest distance from its to final position. And, here the final and initial position is same since he comes to its initial position after covering certain distance. So, displacement is 0.
Answer:
0.056 psi more pressure is exerted by filled coat rack than an empty coat rack.
Explanation:
First we find the pressure exerted by the rack without coat. So, for that purpose, we use formula:
P₁ = F/A
where,
P₁ = Pressure exerted by empty rack = ?
F = Force exerted by empty rack = Weight of Empty Rack = 40 lb
A = Base Area = 452.4 in²
Therefore,
P₁ = 40 lb/452.4 in²
P₁ = 0.088 psi
Now, we calculate the pressure exerted by the rack along with the coat.
P₂ = F/A
where,
P₂ = Pressure exerted by rack filled with coats= ?
F = Force exerted by filled rack = Weight of Filled Rack = 65 lb
A = Base Area = 452.4 in²
Therefore,
P₂ = 65 lb/452.4 in²
P₂ = 0.144 psi
Now, the difference between both pressures is:
ΔP = P₂ - P₁
ΔP = 0.144 psi - 0.088 psi
<u>ΔP = 0.056 psi</u>
<span>v/2
This is an exercise in the conservation of momentum.
The collision specified is a non-elastic collision since the railroad cars didn't bounce away from each other. For the equations, I'll use the following variables.
r1 = momentum of railroad car 1
r2 = momentum of railroad car 2
x = velocity after collision
Prior to the collision, the momentum of the system was
r1 + r2
mv + m*0
So the total momentum is mv
After the collision, both cars move at the same velocity since it was non-elastic, so
r1 + r2
mx + mx
x(m + m)
x(2m)
And since the momentum has to match, we can set the equations equal to each other, so:
x(2m) = mv
x(2) = v
x = v/2
Therefore the speed immediately after collision was v/2</span>
Answer:
ωf = 4.53 rad/s
Explanation:
By conservation of the angular momentum:
Ib*ωb = (Ib + Ic)*ωf
Where
Ib is the inertia of the ball
ωb is the initial angular velocity of the ball
Ic is the inertia of the catcher
ωf is the final angular velocity of the system
We need to calculate first Ib, Ic, ωb:
ωb = Vb / (L/2) = 16 / (1.2/2) = 26.67 m/s
Now, ωf will be:
We are given
damped harmonic oscillation force = k
mass = m
damping constant = b1
amplitude = a1
driving angular frequency = k/m
I think we asked for the amplitude of the force at different damping constant
The formula to use is
A = (F/ (√(k - m w²)² + (b² w²))
Simply substitute and solve for A in terms of a1