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JulijaS [17]
2 years ago
5

The recoil of a cannon is probably familiar to anyone who has watched pirate movies. This is a classic problem in momentum conse

rvation. Consider a wheeled, 500 kg cannon firing a 2 kg cannonball horizontally from a ship. The ball leaves the cannon traveling at 200 m/s. At what speed does the cannon recoil as a result?
Physics
1 answer:
babunello [35]2 years ago
8 0

Answer:

0.8 m/s

Explanation:

From the question,

MV = mv............................. Equation 1

Where M = mass of the cannon, m = mass of the cannonball, V = velocity of the cannon, v = velocity of the cannon ball.

make V the subject of the equation

V = mv/M........................ Equation 2

Given: m = 2 kg, M = 500 kg, v = 200 m/s

Substitute these value into equation 2

V = (2×200)/500

V = 400/500

V = 0.8 m/s

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3 years ago
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What magnification will be produced by a lens of power –4.00 D (such as might be used to correct myopia) if an object is held 43
kiruha [24]

Answer:

The magnification is m  = 0.3674

Explanation:

From the question we are told that

  The  power of the lens is  P = -4.00 D(dioptre)

Generally  1 dioptre = 1 \ meter

  The object distance is u =  -43 \ cm the negative sign is because the distance is measured in the opposite direction of incident light (i.e away )

 Generally the focal length is mathematically represented as

          f = \frac{1}{P}  

   =>f = \frac{1}{4.00 }  

  =>  f = 0.25 \ m

converting to  cm  

 =>   f = 0.25 \ m = 0.25 * 100 = 25 \ cm

Generally from lens equation  we have that  

     \frac{1}{f} +\frac{1}{v} -\frac{1}{u}

=>  \frac{1}{25} +\frac{1}{v} -\frac{1}{-43}

=>   v =  -15.8 \ cm

Generally the magnification is mathematically represented as

      m  = \frac{v}{u}

=>    m  = \frac{- 15.8}{-43}

=>    m  = 0.3674

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2 years ago
Which of these is not a layer of the skin?
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3 years ago
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A satellite orbits the earth a distance of 1.50 × 107 m above the planet's surface and takes 8.65 hours for each revolution abou
kupik [55]

Answer:

The acceleration of the satellite is 0.87 m/s^{2}

Explanation:

The acceleration in a circular motion is defined as:

a = \frac{v^{2}}{r}  (1)

Where a is the centripetal acceleration, v the velocity and r is the radius.

The equation of the orbital velocity is defined as

v = \frac{2 \pi r}{T} (2)

Where r is the radius and T is the period

For this particular case, the radius will be the sum of the high of the satellite (1.50x10^{7} m) and the Earth radius (6.38x10^{6} m) :

r = 1.50x10^{7} m+6.38x10^{6}m

r = 21.38x10^{6}m

Then, equation 2 can be used:

T = 8.65 hrs \cdot \frac{3600 s}{1hrs} ⇒ 31140 s

v = \frac{2 \pi (21.38x10^{6}m)}{31140s}

v = 4313 m/s

Finally equation 1 can be used:

a = \frac{(4313m/s)^{2}}{21.38x10^{6}m}    

a = 0.87 m/s^{2}

Hence, the acceleration of the satellite is 0.87 m/s^{2}

6 0
3 years ago
The concentration of an acid or base refers to how completely it dissociates in
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Answer:

False

Explanation:

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