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Inessa [10]
3 years ago
6

An iron cannon ball and a bowling ball are dropped at the same time from the top of a building. At the instant before the balls

hit the sidewalk, the heavier cannonball has a greater?A. velocity. B. aceleration. C. kinetic energy. D. all of these are the same for the two balls.
Physics
1 answer:
scoundrel [369]3 years ago
6 0

Answer:

<em>C. kinetic energy</em>

Explanation:

<u>Free Fall </u>

When objects are dropped in free air (no friction), they are attracted to the Earth's surface by the force of gravity. Objects start from zero speed and gradually increase it because of the effect of the acceleration of gravity whose value can be considered as constant g=9.8\ m/s^2.

The final speed of the object at time t is

v_f=g.t

Note that the final speed does not depend on the mass of the object. The kinetic energy is

\displaystyle k=\frac{mv_f^2}{2}

The kinetic energy depends on the speed and the mass, so heavier objects dropped from the same height will have more kinetic energy. Let's analyze the options given in the question .

A. As shown above, the speed (or magnitude of the velocity) is the same regardless of the object's mass, so the heavier cannonball and the iron cannon will reach the ground at the same speed. Incorrect option

B. For every falling object in free air, the only acceleration is the acceleration of gravity. This option is not correct either

C. The kinetic energy is greater for the heavier cannonball because it has more mass, as discussed above. Correct option

D. Only the C. option is correct. This is not

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Read 2 more answers
Two stones are launched from the top of a tall building. One stoneis thrown in a direction 30.0^\circ above the horizontal with
Butoxors [25]

Answer:

Part A)

t(1) > t(2), the stone thrown 30 above the horizontal spends more time in the air.

Part B)

x(f1) > x(f2), the first stone will land farther away from the building.

Explanation:

<u>Part A)</u>

Let's use the parabolic motion equation to solve it. Let's define the variables:

  • y(i) is the initial height, it is a constant.
  • y(f) is the final height, in our case is 0
  • v(i) is the initial velocity (v(i)=16 m/s)
  • θ1 is the first angle, 30°
  • θ2 is the first angle, -30°

For the first stone

y_{f1}=y_{i1}+v*sin(\theta_{1})t_{1}-0.5gt_{1}^{2}              

0=y_{i1}+16*sin(30)t_{1}-0.5*9.81*t_{1}^{2}

0=y_{i1}+8t_{1}-4.905*t_{1}^{2} (1)  

For the second stone  

0=y_{i2}+16*sin(-30)t_{2}-4.905t_{2}^{2}    

0=y_{i2}-8t_{2}-4.905t_{2}^{2} (2)            

 

If we solve the equation (1) we will have:

t_{1}=\frac{-8\pm \sqrt{64+19.62*y_{i}}}{-9.81}  

We can do the same procedure for the equation (2)

t_{1}=\frac{8\pm \sqrt{64+19.62*y_{i}}}{-9.81}

We can analyze each solution to see which one spends more time in the air.

It is easy to see that the value inside the square root of each equation is always greater than 8, assuming that the height of the building is > 0. Now, to get positive values of t(1) and t(2) we need to take the negative option of the square root.

Therefore, t(1) > t(2), it means that the stone thrown 30 above the horizontal spends more time in the air.

<u>Part B)</u>

We can use the equation of the horizontal position here.

<u>First stone</u>

x_{f1}=x_{i1}+vcos(30)t_{1}

x_{f1}=0+13.86*t_{1}

x_{f1}=13.86*t_{1}

<u>Second stone</u>

x_{2}=x_{i2}+vcos(-30)t_{2}

x_{1}=0+13.86*t_{1}

x_{1}=13.86*t_{2}

Knowing that t(1) > t(2) then x(f1) > x(f2)

Therefore, the first stone will land farther away from the building.

They land at different points at different times.

I hope it helps you!

3 0
3 years ago
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