When both particles, the electron and the proton move at the same speed, they may have differences with their de Broglie wavelength, the particle that would have a longer wavelength would be the proton since the wavelength is in direct proportionality with the mass of the particle.
<span>The 2nd truck was overloaded with a load of 16833 kg instead of the permissible load of 8000 kg.
The key here is the conservation of momentum.
For the first truck, the momentum is
0(5100 + 4300)
The second truck has a starting momentum of
60(5100 + x)
And finally, after the collision, the momentum of the whole system is
42(5100 + 4300 + 5100 + x)
So let's set the equations for before and after the collision equal to each other.
0(5100 + 4300) + 60(5100 + x) = 42(5100 + 4300 + 5100 + x)
And solve for x, first by adding the constant terms
0(5100 + 4300) + 60(5100 + x) = 42(14500 + x)
Getting rid of the zero term
60(5100 + x) = 42(14500 + x)
Distribute the 60 and the 42.
60*5100 + 60x = 42*14500 + 42x
306000 + 60x = 609000 + 42x
Subtract 42x from both sides
306000 + 18x = 609000
Subtract 306000 from both sides
18x = 303000
And divide both sides by 18
x = 16833.33
So we have the 2nd truck with a load of 16833.33 kg, which is well over it's maximum permissible load of 8000 kg. Let's verify the results by plugging that mass into the before and after collision momentums.
60(5100 + 16833.33) = 60(21933.33) = 1316000
42(5100 + 4300 + 5100 + 16833.33) = 42(31333.33) = 1316000
They match. The 2nd truck was definitely over loaded.</span>
Answer:
a) 2.87 m/s
b) 3.23 m/s
Explanation:
The avergare velocity can be found dividing the length traveled d by the total time t.
a)
For the first part we easily know the total traveled length which is:
d = 50.2 m + 50.2 m = 100.4 m
The time can be found dividing the distance by the velocity:
t1 = 50.2 m / 2.21 m/s = 22.7149 s
t2 = 50.2 m / 4.11 m/s = 12.2141 s
t = t1 +t2 = 34.9290 s
Therefore, the average velocity is:
v = d/t =2.87 m/s
b)
Here we can easily know the total time:
t = 1 min + 1.16 min = 129.6 s
Now the distance wil be found multiplying each velocity by the time it has travelled:
d1 = 2.21 m/s * 60 s = 132.6 m
d2 = 4.11 m/s *(1.16 * 60 s) = 286.056 m
d = 418.656 m
Therefore, the average velocity is:
v = d/t =3.23 m/s
Answer:
A) d_o = 20.7 cm
B) h_i = 1.014 m
Explanation:
A) To solve this, we will use the lens equation formula;
1/f = 1/d_o + 1/d_i
Where;
f is focal Length = 20 cm = 0.2
d_o is object distance
d_i is image distance = 6m
1/0.2 = 1/d_o + 1/6
1/d_o = 1/0.2 - 1/6
1/d_o = 4.8333
d_o = 1/4.8333
d_o = 0.207 m
d_o = 20.7 cm
B) to solve this, we will use the magnification equation;
M = h_i/h_o = d_i/d_o
Where;
h_o = 3.5 cm = 0.035 m
d_i = 6 m
d_o = 20.7 cm = 0.207 m
Thus;
h_i = (6/0.207) × 0.035
h_i = 1.014 m
