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frosja888 [35]
3 years ago
9

A column of soldiers, marching at 114 steps per minute, keep in step with the beat of a drummer at the head of the column. It is

observed that the soldiers in the rear end of the column are striding forward with the left foot when the drummer is advancing with the right. What is the approximate length of the column? (Take the speed of sound to be 343 m/s.)
Physics
1 answer:
kipiarov [429]3 years ago
5 0

Answer:

length of column is 140.42 m

Explanation:

in 1 min the soldiers travels 114 steps

so to complete 1 step the time taken by the soldiers =

t =  \dfrac{1\ min}{114}

t =  \dfrac{60}{114}  

t = 0.526 s

speed of sound = 343 m/s

length of the column = v × t

                                   = 343 m/s × 0.526 s

                                   = 140.42 m

hence the length of column is 140.42 m

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The pressure of liquid column is given by p=hpg, where h is depth, p is density and g is acceleration due to gravity.

Therefore, pressure of the liquid column increases with depth. The height of the blood column in a human body is more at feet than at the brain. Therefore, the blood pressure in humans is greater at the feet than the brain.

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Explanation:

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6 0
3 years ago
g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart
Veseljchak [2.6K]

Answer:

The speed of q₂ is 4\sqrt{10}\ m/s

Explanation:

Given that,

Distance = 0.4 m apart

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

We need to calculate the speed of q₂

Using conservation of energy

E_{i}=E_{f}

\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2

\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})

Put the value into the formula

\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{f}^2)=9\times10^{9}\times-2\times10^{-6}\times-8\times10^{-6}(\dfrac{1}{(0.4)}-\dfrac{1}{(0.8)})

0.00075(400-v_{f}^2)=0.18


400-v_{f}^2=\dfrac{0.18}{0.00075}

-v_{f}^2=240-400

v_{f}^2=160

v_{f}=4\sqrt{10}\ m/s

Hence, The speed of q₂ is 4\sqrt{10}\ m/s

7 0
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If two parallel conductors are both free to move and are carrying current in the same direction, what would happen if the curren
pishuonlain [190]

The wires would remain attracted to each other.

Option D.

Explanation:

It is known that magnetic flux will be generated in conductors with varying emf. So when current is flowing in two parallel conductors, the magnetic flux will be generated in those wires. If the current is flowing in same direction in both the wires, then the magnetic flux will be generated towards inside and outside the wires. Thus, the wire will get attracted to each other till the time the current is flowing in the same direction in both the wires. So if the current flow in each wire was reversed at the same time, then the wire would remain attracted to each other.

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