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alexandr402 [8]
3 years ago
10

As the army approached the bridge into the city, the commander told them to break step. He did not want the soldiers to march in

cadence. Why did the commander give this order?
Physics
2 answers:
GrogVix [38]3 years ago
5 0

Answer:

Their rhythmic marching might resonate through the bridge and start the bridge swaying.

HOPE THIS HELPS

Aleks [24]3 years ago
3 0

At a definite point, the bridge would begin oscillating to the matching rhythm as that of the marching footsteps. This oscillation would touch a determined peak when the bridge can no longer tolerate its own power and later collapses. So, soldiers are systematic to break their steps while passing a bridge.

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A uniform rod of length L, mass M, is suspended by two thin strings. Which of the following statements is true
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3 0
2 years ago
What fraction of 5 MeV α particles will be scattered through angles greater than 8.5° from a gold foil (Z = 79, density = 19.3 g
aalyn [17]

Answer:

f(8) = \pi (5.90x10^{28} atoms/m^3) *(10^{-8} m) (\frac{2*79*(1.6x10^{-19}C)^2}{8 \pi (8.85 x10^{-12} C^2/Nm^2)*(5x10^6 Nm) *1.6x10^{-19}C})* cot^2 (8/2)

And after do all the operations we got:

f(8) = 1.96 x10^{-4] atoms/m^2

And that would be the final answer for this case.

Explanation:

For this case we can use the fomrula for the fraction of incident particles scattered by an angle \theta, given by:

f(\theta) = \pi nt (\frac{Z_1 e Z_2 e}{8 \pi e_o K})^2 cos^2 (\theta/2)

Where:

Z_1 e represent the charge of the projectile (Z1=2)

Z_2 e is the target charge (z2=79)

K= 5x10^6 eV represent the kinetic energy of incident particle

n represent the density of target particles (we need to find it first)

t= 10^{-8] m represent the thicknss of the foil

The first step would be calculate the density of target particles with the following formula:

n =\frac{\rho N_A N_M}{M_g}

Where:

\rho = 19.3 g/m^3= 19300 Kg/m^3

N_A = 6.022 x10^{23} molecules/mol the Avogadro's number

N_M = 1 represent the atoms per molecule

M_g = 197 g/mol = 0.197 Kg/mol represent the molecular weigth

If we replace we got:

n = \frac{19300 kg/m^3 *6.022x10^{23} molecu/mol * 1 atom/mole}{0.197 Kg/mol}= 5.90 x106{28] atoms/m^3

Now we can calculate the fraction of 5 MeV alpha particles that would be scatteres with angle higher than 8 degrees in a piece of thickness t=10^{-8}m

And using the first formula we got:

f(8) = \pi (5.90x10^{28} atoms/m^3) *(10^{-8} m) (\frac{2*79*(1.6x10^{-19}C)^2}{8 \pi (8.85 x10^{-12} C^2/Nm^2)*(5x10^6 Nm) *1.6x10^{-19}C})* cot^2 (8/2)

And after do all the operations we got:

f(8) = 1.96 x10^{-4] atoms/m^2

And that would be the final answer for this case.

4 0
3 years ago
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