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astra-53 [7]
3 years ago
13

A 500.0-kilogram cannon is at rest on a frictionless surface. A remote triggering device causes a 5.0-kilogram projectile to be

fired from the cannon moving at 30 m/s. What is the recoil speed of the cannon?
A. 83.3 m/s
B. 0.3 m/s
C. 16.7 m/s
D. 75 m/s
Physics
1 answer:
KiRa [710]3 years ago
3 0
From the law of conservation of momentum the initial momentum is equivalent to the final momentum. Therefore, the total momentum before firing will be equivalent to the momentum after firing. if we take mass of the projectile to be mass i and that of the cannon as mass ii.
The initial momentum will be given by m1v1 +m2v1 
This initial momentum is equivalent to zero since initially both the cannon and the projectile are at zero velocity.
The final momentum is given by m1v2 + m2v3, where v2 is the velocity of the projectile  and v3 is the recoil velocity of the cannon.
Therefore, m1v2 + m2v3 =0
              ( 5 × 30 ) + ( 500 × v3) =0
               500 v3 = -150
                       v3 =  - 0.3 m/s
The velocity is negative indicating that the cannon moved in the opposite direction. therefore, the recoil velocity is 0.3 m/s
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balls a, with a mass of 20 kg, is moving to the right at 20 m/s. At what velocity should Ball B, with a mass of 40 kg, move so t
Leya [2.2K]
In that case, their momentum must be equal. 
So, m1v1 = m2v2
20 * 20 = 40 * v2
v2 = 400 / 40
v2 = 10

In short, Your Answer would be: 10 m/s

Hope this helps!
3 0
3 years ago
If the column of mercury in a barometer stands at 72.6 cm, what is the atmospheric pressure?
lakkis [162]

Answer:pressure = density * acceleration due to gravity * height

H=72.6cm= 0.726m

P=0.726*13.6*10^3*9.8

P=96761.28Pa

Explanation:

6 0
3 years ago
An object of mass m = 5.0 kg hangs from a cord around a light pulley: The length of the cord between the oscillator and the pull
puteri [66]

Answer:

\mu=0.0049Kg/m

Explanation:

When a standing wave is formed with six loops means the normal mode of the wave is n=6, the frequency of the normal mode is given by the expression:

f_n=\frac{nv}{2L}

Where L is the length of the string and v the velocity of propagation. Use this expression to find the value of v.

f_6=\frac{6v}{2L}\\(150)=\frac{6v}{2(2)} \\150=\frac{3v}{2} \\3v=150(2)\\ v=\frac{300}{3} \\v=100m/s

The velocity of propagation is given by the expression:

v=\sqrt{\frac{T}{\mu }

Where \mu is the desirable variable of the problem, the linear mass density, and T is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord:

T=W=mg=(5)(9.81)=49.05N

With the value of the tension and the velocity you can find the mass density:

v=\sqrt{\frac{T}{\mu}

v^2=\frac{T}{\mu}\\ \mu=\frac{T}{v^2} =\frac{49.05}{(100)^2} =\frac{49.05}{10000} =0.0049Kg/m

6 0
3 years ago
A force of 150N at an angle of 60 degree to the horizontal to pull a box through a distance of 50m calculate the work done
Alexxandr [17]
  • Force=150N
  • Angle=60°
  • Displacement=50m

\boxed{\sf W=Fscos\Theta}

\\ \sf\longmapsto W=150(50)cos 60

\\ \sf\longmapsto W=7500\times \dfrac{1}{2}

\\ \sf\longmapsto W=3750J

6 0
3 years ago
Help pls, see picture. Will mark Brainliest
Pani-rosa [81]

Answer:

a ) option 2 is correct

b) -ve acceleration for upward motion ,0 acceleration at top point ,+ve acceleration on downward motion ...

Explanation:

mark me as brainliest ❤️

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3 years ago
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