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sukhopar [10]
3 years ago
5

A column of soldiers, marching at 121 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:
Novay_Z [31]3 years ago
7 0

There’s a first “clump”/drum beat every half second. That clump will travel about 170m in half a second. Someone 170m away would do their “clump” as the second “clump” was taking place. I think. <span> </span>

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If a car driving down the road doubles its speed what happens to its kinetic energy?
ivanzaharov [21]

Answer:

Kinetic energy doubles

7 0
2 years ago
He shoots the tree stump, which has a mass of (M), with a bullet of mass (m) traveling at some velocity (vbullet), and the bulle
Ymorist [56]

Answer:

Explanation:

Given

mass of tree stump is M

mass bullet is m

velocity of bullet is v

Conserving momentum for bullet and tree stump

Initial Momentum P_i=mv

Suppose v_0 is the velocity of the system

Final Momentum P_f=(M+m)v_0

Initial momentum =Final Momentum

mv=(m+M)v_0

v_0=\frac{mv}{m+M}

4 0
3 years ago
Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its
ruslelena [56]

Answer:

a

The orbital speed is v= 2.6*10^{3} m/s

b

The escape velocity of the rocket is  v_e= 3.72 *10^3 m/s

Explanation:

Generally angular velocity is mathematically represented as

            w = \frac{2 \pi}{T}

Where T is the period which is given as 1.6 days = 1.6 *24 *60*60 = 138240 sec

       Substituting the value

         w = \frac{2 \pi}{138240}

             = 4.54*10^ {-5} rad /sec

At the point when the rocket is on a circular orbit  

   The gravitational force =  centripetal force and this can be mathematically represented as

              \frac{GMm}{r^2} = mr w^2

Where  G is the universal gravitational constant with a value  G = 6.67*10^{-11}

            M is the mass of the earth with a constant value of M = 5.98*10^{24}kg

            r is the distance between earth and circular orbit where the rocke is found

               Making r the subject

                     r = \sqrt[3]{\frac{GM}{w^2} }

                        = \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }

                        = 5.78 *10^7 m

The orbital speed is represented mathematically as

                   v=wr

Substituting value

                  v= (5.78*10^7)(4.54*10^{-5})

                     v= 2.6*10^{3} m/s    

The escape velocity is mathematically represented as

                            v_e = \sqrt{\frac{2GM}{r} }

Substituting values

                             = \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }

                             v_e= 3.72 *10^3 m/s

7 0
3 years ago
If the particle has mass m, how fast must it be moving away from the Sun's center of mass to escape the gravitational influence
IgorLugansk [536]

Answer:

Explanation:

M = 1.989 x 10^30 kg

R = 6.96 x 10^8 m

G = 6.67 x 10^-11 Nm²/kg²

Let the velocity is v.

v=\sqrt{\frac{2GM}{R}}

v=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 1.989\times 10^{30}}{6.96\times 10^{8}}}

v = 6.17 x 10^5 m/s

3 0
3 years ago
A light flashes at position x=0m. One microsecond later, a light flashes at position x=1000m. In a second reference frame, movin
padilas [110]

Answer:

To the right relative to the original frame.

Explanation:

In first reference frame <em>S</em>,

Spatial interval of the event, \rm \Delta x=1000\ m-0\ m=1000\ m.

Temporal interval of the event, \rm \Delta t = 1\ \mu s=10^{-6}\ s.

In the second reference frame <em>S'</em>, the two flashes are simultaneous, which means that the temporal interval of the event in this frame is \rm \Delta t'=0\ s.

The speed of the frame <em>S' </em>with respect to frame <em>S</em> = v.

According to the Lorentz transformation,

\rm \Delta t'=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\left( \Delta t-\dfrac{v\Delta x}{c^2}\right ).\\\\Since,\ \Delta t'=0,\\\therefore \dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\left( \Delta t-\dfrac{v\Delta x}{c^2}\right )=0\\\Rightarrow \Delta t-\dfrac{v\Delta x}{c^2}=0\\\dfrac{v\Delta x}{c^2}=\Delta t\\v=\dfrac{c^2\Delta t}{\Delta x }.\\\\Also, \ \Delta t,\ \Delta x>0\ \Rightarrow v>0.

And positive v means the velocity of the second frame<em> </em><em>S'</em> is along the positive x-axis direction, i.e., to the right direction relative to the original frame <em>S</em>.

7 0
3 years ago
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