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enyata [817]
3 years ago
5

A revolutionary war cannon, with a mass of 2240 kg, fires a 15.5 kg ball horizontally. The cannonball has a speed of 131 m/s aft

er it has left the barrel. The cannon carriage is on a flat platform and is free to roll horizontally. What is the speed of the cannon immediately after it was fired?Answer in units of m/s.
Physics
1 answer:
My name is Ann [436]3 years ago
8 0

Answer:

v' = -0.0906 m/s

Explanation:

given,

mass of cannon, M = 2240 Kg

mass of the ball, m = 15.5 Kg

speed of cannon ball, v = 131 m/s

speed of he cannon = ?

initial speed of cannon and the cannon ball is equal to 0 m/s

using conservation of energy

(M+m)V = M v' + m v

(M+m) x 0= 22400 v' + 15.5 x 131

22400 v' = -2030.5

v' = -0.0906 m/s

negative sign represent the canon will move in opposite direction of the ball.

hence, speed of cannon is equal to 0.0906 m/s

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A model rocket is launched directly upward at a speed of 16 meters per second from a height of 2 meters. The function f(t)=−4.9t
Crank

Answer:

Hmax=15.06 meters

Explanation:The question ask for the maximum value of the function f(t) which can be find by find the maxima of the function

The maxima of the function occurs when the slope is zero. i.e.

\frac{df}{dt} =0\\\frac{df}{dt} =\frac{d}{dt} (-4.9t^2+16t+2)\\\frac{df}{dt} =-4.9*2t+16\\-9.8t+16=0\\t=16/9.8\\t=1.63 secs

Hence the maxima occurs at t=1.63 seconds

The maximum value of f is

f(1.63)=-4.9(1.63^2)+16(1.63)+2\\f(1.63)=15.06\\

hence maximum height is found to be

Hmax=15.06 meters

8 0
3 years ago
Three joggers are running along straight lines as follows: Jogger A, with a mass of 55.2kg, is traveling along the line y=6.00m
frutty [35]

Answer:

L = - 1361.591 k Kgm/s

Explanation:

Given

mA = 55.2 Kg

vA = 3.45 m/s

rA = 6.00 m

mB = 62.4 Kg

vB = 4.23 m/s

rB = 3.00 m

mC = 72.1 Kg

vC = 4.75 m/s

rC = - 5.00 m

then we apply the equation

L =  (mv x r)

⇒  LA = mA*vA x rA = 55.2 *(3.45 i)x(6 j) = (1142.64 k) Kgm/s

⇒  LB = mB*vB x rB = 62.4 *(4.23 j)x(3 i) = (- 791.856 k) Kgm/s

⇒  LC = mC*vC x rC = 72.1 *(- 4.75 j)x(- 5 i) = (- 1712.375 k) Kgm/s

Finally, the total counterclockwise angular momentum of the three joggers about the origin is

L = LA + LB + LC = (1142.64 - 791.856 -1712.375) k  Kgm/s

L = - 1361.591 k Kgm/s

7 0
3 years ago
You travel 85 miles in 2 hours, you then travel 75 miles in 1.5 hours. After a 0.5 hour rest break you then travel 55 miles in 1
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Total 215 miles in 5 hours .hope it helps
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3 years ago
b. Comparing and Contrasting Compare the change in atmospheric pressure with elevation to the change in water pressure
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Answer:

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3 years ago
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Newton’s laws do not apply to small objects?
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Answer:

Yes Newton's laws apply to small objects

EX: Newton s first law

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or body at motion always want to be at motion

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if so e apply some force then it comes motion

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