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adelina 88 [10]
3 years ago
12

A Cannonball is shot at an angle of 35.0 degrees and is in flight for 11.0 seconds before hitting the ground at the same height

from which it was shot.
A. What is the magnitude of the inital velocity?B. What was the maximum height reached by the cannonball?C. How far, horizontally, did it travel?
Physics
1 answer:
nikdorinn [45]3 years ago
4 0

Answer:

Explanation:

According to Equations of Projectile motion :

Time\ of\ Flight = \frac{2vsin(x)}{g}

vsin(x) = 11 * 9.8 / 2 = 53.9 m/sec

(A) v (Initial velocity) = 11 * 9.8 / 2 * sin(35) = 94.56 m/sec

Maximum Height = \frac{(vsinx)^{2} }{2g}

(B) Maximum Height = 53.9 * 53.9 / 2 * 9.8 = 142.2 m

Horizontal Range = vcosx * t

(C) Horizontal Range = 94.56 * 0.81 * 11 = 842.52 m

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oee [108]

The 3,402 has units of kg-m/s.  That's momentum.  The 20,000 has units of Newtons.  That's force.  Momentum and force are different physical things, and you can't convert them from one to the other.

The best I can do for you is something like this:

Let's say you have a moving object with 3,402 kg-m/s of momentum, and you want to STOP it completely.  You want to stand in front of it and push back on it, hard enough and for long enough to CHANGE its momentum from 3,402 kg-m/s to zero.

Also ... there's a limit to how hard you can push.  The most force you can exert is 20,000 Newtons.

The amount you'll change its momentum is called the <u><em>impulse</em></u> you give it.  The quantity of impulse is (force) x (length of time you push on it).

So you need to keep pushing it back for (T seconds) long enough so that

(20,000 Newtons of force) x (T seconds) = 3,402 kg-m/s of momentum .

Divide each side of that equation by (20,000 Newtons). Then it says:

(T seconds) = (3,402 kg-m/s) / (20,000 Newtons)

<em>T = 0.1701 second</em>

And that's how you provide just enough impulse to stop the flying object ... push on it with 20,000 Newtons of force for exactly 0.1701 second, and it loses all its momentum and falls out of the air onto the ground at your feet.

This story is the closest I can come to anything that looks like "convert"ing momentum into force.

3 0
3 years ago
Match the words in the left-hand column to the appropriate blank in the sentences in the right-hand column. Use each word only o
Vlad [161]

Answer:

1. About a trillion comets are thought to be located far, far beyond Pluto in the Oort cloud.

2. The bright spherical part of a comet observed when it is close to the Sun is the coma .

3. A comet's plasma tail stretches directly away from the Sun.

4. A comet's nucleus is the frozen portion of a comet.

5. Particles ejected from a comet can cause a(n) meteor shower on Earth.

6. The Kuiper belt extends from about beyond the orbit of Neptune to about twice the distance of Neptune from the Sun.  

Explanation:

1. Oort cloud is a theoretical cloud of icy objects in interstellar space. Its inner limit begins about 2000-5000 AU from sun outer limit (about 100000-200000 AU) defines the boundary of solar system.  

Note: AU is the astronomical unit which is defined as the distance between the Sun and Earth.  

2. Comets are described as “dirty snowballs” which is a mixture of frozen gases and dust. When this snowball gets closer to the Sun, gas evaporates, and a fuzzy haze of gas and dust emerges. This fuzzy tail is the coma.

3. Dust and gas in the coma interact with the sunlight and the solar wind. Charged particles in the coma are caught up in the magnetic field of sun and flow away from the sun.  

4. Nucleus is the dirty snowball which consists of rock, dust and frozen gas.  

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3 years ago
A bullet of mass 11 g strikes a ballistic pendulum of mass 1.9 kg. The center of mass of the pendulum rises a vertical distance
Ymorist [56]

Answer:

The bullet's initial speed is 243.21 m/s.

Explanation:

Given that,

Mass of the bullet, m_b=11\ g=0.011\ kg

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The center of mass of the pendulum rises a vertical distance of 10 cm.

We need to find the bullet's initial speed if it is assumed that the bullet remains embedded in the pendulum. Let it is v. In this case, the energy of the system remains conserved. The kinetic energy of the bullet gets converted to potential energy for the whole system. So,

\dfrac{1}{2}(m_b+m_p)V^2 =(m_b+m_p)gh\\\\V=\sqrt{2gh} \ .................(1)

V is the speed of the bullet and pendulum at the time of collision

Now using conservation of momentum as :

m_bv=(m_p+m_b)V

Put the value of V from equation (1) in above equation as :

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So, the bullet's initial speed is 243.21 m/s.

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