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deff fn [24]
4 years ago
10

Sir Dudley the Dextrous is loading cannons on top of the castle's tallest towers in preparation for an expected attack. He slips

, and a cannonball of mass 14.0 kg 14.0 kg rolls to the edge of the tower, which has a perfectly horizontal surface, and falls to the ground below. It takes the cannonball 3.15 s 3.15 s to fall to the ground, and air resistance is negligible. How much work is done by gravity on the falling cannonball?
Physics
1 answer:
JulsSmile [24]4 years ago
4 0

Answer:

6667.9 J

Explanation:

We are given that

Mass of cannon.m=14 kg

Time,t=3.15 s

Initial velocity,u=0

We have to find the work done by gravity on the falling cannon ball.

g=9.8 m/s^2

h=ut+\frac{1}{2}gt^2

Substitute the values

h=0+\frac{1}{2}\times 9.8\times (3.15)^2=48.6 m

Work done,W=mgh

Using the formula

W=14\times 9.8\times 48.6=6667.9 J

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Answer:

t=0.42s

Explanation:

Here you have an inelastic collision. By the conservation of the momentum you have:

m_1v_1+m_2v_2=(m_1+m_2)v

m1: mass of the bullet

m2: wooden block mass

v1: velocity of the bullet

v2: velocity of the wooden block

v: velocity of bullet and wooden block after the collision.

By noticing that after the collision, both objects reach the same height from where the wooden block was dropped, you can assume that v is equal to the negative of v2. In other words:

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Where you assumed that the negative direction is upward. By replacing and doing v2 the subject of the formula you get:

-(0.0129kg)(767m/s)+(1.17kg)v_2=(1.1829kg)(-v_2)\\\\v_2=4.20m/s

Now, with this information you can use the equation for the final speed of an accelerated motion and doing t the subject of the formula. IN other words:

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