Answer:
F₁ / F₂ = 10
therefore the first out is 10 times greater than the second barrier
Explanation:
For this exercise let's use the relationship between momentum and momentum.
I = F t = Δp
in this case the final velocity is zero
F t = 0 -m v₀
F = m v₀ / t
in order to answer the question we must assume that the two vehicles have the same mass and speed
concrete barrier
F₁ = -p₀ / 0.1
F₁ = - 10 p₀
barrier collapses
F₂ = -p₀ / 1
let's look for the relationship of the forces
F₁ / F₂ = 10
therefore the first out is 10 times greater than the second barrier
Answer:
B) Power
Explanation:
The power is defined by the following equation:
P = W / t
where:
W = work = Force * Distance = [Newton] * [meter]
t = time = seconds
The units for work are give en Newton per second, which is equal to Joules
And for power the unit used commonly is Watts, therefore:
Watts = (Joule/second)
Answer:
6.5454 m
Explanation:
Let the distance from the wire carrying 3 A current is x
Then the distance from the the carrying current 8 A is 24-x
We know that magnetic field due to long wire is given by 
It is given that magnetic field is zero at some distance so
Here 
So 
Answer:
Explanation:
If friction is neglected, the wheel cannot roll and can only slide frictionlessly and will have the same velocity at the bottom of the ramp as if it had been in free fall as it has converted the same amount of potential energy.
mgh = ½mv²
v = √(2gh) = √(2(9.81)(2.00)) = 6.26418... = 6.26 m/s
However if we do not ignore all friction and the wheel rolls without slipping down the slope, the potential energy becomes linear and rotational kinetic energy
mgh = ½mv² + ½Iω²
mgh = ½mv² + ½(½mR²)(v/R)²
2gh = v² + ½v²
2gh = 3v²/2
v = √(4gh/3) =√(4(9.81)(2.00)/3) = 5.11468... = 5.11 m/s
Answer:
he peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.
Explanation:
In a resonance experiment, the amplitude of the system is plotted as a function of the frequency, finding maximums for the values where some natural frequency of the system coincides with the excitation frequency.
In a Fourier transform spectrum, the amplitude of the frequencies present is the signal, whereby each peak corresponds to a natural frequency of the system.
From this explanation we can see that in the first case the peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.
