Answer:
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The velocity of the cannonball is 150 m/s, the right option is B. 150 m/s.
The question can be solved, using Newton's second law of motion.
Note: Momentum of the cannon = momentum of the cannonball.
<h3>
Formula:</h3>
- MV = mv................. Equation 1
<h3>Where:</h3>
- M = mass of the cannon
- m = mass of the cannonball
- V = velocity of the cannon
- v = velocity of the cannonball
Make v the subject of the equation.
- v = MV/m................ Equation 2
From the question,
<h3>Given: </h3>
- M = 500 kg
- V = 3 m/s
- m = 10 kg.
Substitute these values into equation 2.
- v = (500×3)/10
- v = 150 m/s.
Hence, The velocity of the cannonball is 150 m/s, the right option is B. 150 m/s.
Learn more about Newton's second law here: brainly.com/question/25545050
Answer:
The graph appears to be in error.
The actual figure appears to be a rhombus with sides of 5 and 15 with a height of 5
The work done (F * S) is the area of the rhombus
1/2 * (5 +15) * 5 = 50 J
Answer:
a) 19.4 m/s
b) 19 m/s
Explanation:
a) In the given question,
the potential energy at the initial point = Ui = 0
the potential energy at the final point = Uf = mgh
the kinetic energy at the initial point = Ki = 1/2 mv₀².
the kinetic energy at the final point = Kf = 0
work done by air= Ea= fh = 0.262 N
Now, using the law of conservation of energy
initial energy= final energy
Ki +Ui = Kf + Uf +Ea
1/2 mv₀² + 0 = 0 + mgh + fh
1/2 mv₀² = mgh + fh
h = v₀²/ 2g (1 +f/w)
calculate m
m= w/g = 5.29 /9.8
= 0.54 kg
h = 20 ²/ (2 x9.80) x (1 0.265/5.29)
h = 19.4 m.
b) 1/2 mv² + 2fh = 1/2 mv₀²
Vg = 19 m/s
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