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sergiy2304 [10]
3 years ago
8

A 14 kg cannon ball is fired from a 622 kg brass cannon on a frictionless, horizontal surface at an angle of 15º with respect to

the horizontal. The cannon is constrained to move only in the horizontal direction. The initial velocity of the cannon ball is vi = 32 m/s. 1) Previous to launch, what is the net momentum of the ball and cannon system?
Physics
1 answer:
irinina [24]3 years ago
4 0

Answer:

The net momentum of the system is zero.

Solution:

As per the question:

Mass of cannon ball, M = 14 kg

Mass of the cannon, m = 622 kg

Angle of inclination, \theta = 15^{\circ}

Initial velocity of the cannon, v_{i} = 32 m/s

Now,

In order to find the net momentum of the cannon and cannon ball system before the launch, we apply the principle of conservation of momentum to system:

P_{in, cannon} + P_{in, ball} = P_{final, cannon} + P_{final, ball}

Initially, the momentum of the system is zero.

Also, the velocity of the cannon ball in the horizontal direction equals the velocity of the brass cannon:

v_{cannon} = v_{in, ball}cos\theta

v_{cannon} = 32cos15 = 30.91 m/s

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The world speed record on water was set on October 8, 1978 by Ken Warby. If Ken drove his motorboat a distance of 1500 m in 8.10
maxonik [38]

Answer:

317.52 mi/hr

Explanation:

First convert Meters into miles as the answer is required in miles/ h

1000m = 0.62 mi

Now, convert second into hours

7.45s = 0.0001 hr

The speed of the boat would be

v = 0.62/0.0001

=317.52 mi/hr

3 0
3 years ago
A golfer hits her tee-shot due north towards the fairway. Her shot has an initial velocity of 60 m/s. A 15 m/s wind is blowing i
vovangra [49]

Answer:

c=71.4m/s

\theta=8.54\textdegree

Explanation:

From the question we are told that

Initial velocity of 60 m/s

Wind speed V_w= 15 m/s \angle  45 \textdegree

Generally Resolving vector mathematically

  sin(45\textdegree)15=10.6\\cos(45\textdegree)15=10.6

Generally the equation Pythagoras theorem is given mathematically by

c^2=a^2+b^2

c^2=10.6^2 +(10.6+60)^2

c=\sqrt{10.6^2 +(10.6+60)^2}

Therefore Resultant velocity (m/s)

c=71.4m/s

b)Resultant direction

Generally the equation for solving Resultant direction

\theta=tan^-1(\frac{y}{x})

Therefore

\theta=tan^-1(\frac{10.6}{70.6})

\theta=8.54\textdegree

7 0
3 years ago
An object has an average acceleration of +6.18 m/s2 for 0.266 s . At the end of this time the object's velocity is +9.90 m/s .
Archy [21]

At the start of the 0.266 s, the object's speed was 8.26 m/s.

The question can only be talking about speed, not velocity.

4 0
3 years ago
Why does the temperature decreases at higher altitudes
sergey [27]

temperature decreases at higher altitudes because as air rises the pressure decreases.

3 0
3 years ago
40 POINTS EASY
11111nata11111 [884]
We know, Mechanical Energy = K.E. + P.E.
As ball is at ground, P.E. would be zero. But as it is in motion, it must have some K.E. and that is:

K.E. = 1/2 mv²
K.E. = 1/2 * 1 * 2²
K.E. = 4/2
K.E. = 2 J

In short, Your Answer would be Option B

Hope this helps!
7 0
3 years ago
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