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Readme [11.4K]
4 years ago
11

SOMEONE PRETTY PLEASE FILL THIS OUT Its the ball drop project Lab Report Title: Formulate Hypothesis: What are your independent,

dependent and constant variable? Data and Observations Temperature Height Dropped Bounce Height Ball 1 Ball 2 Ball 3 Temperature Height Dropped Bounce Height Ball 1 : Ball 2: Ball 3: Discussion and Conclusions Type your discussion and conclusions here.
Physics
1 answer:
Tatiana [17]4 years ago
5 0
Please help me tooooornrjjfjenr
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Calculate the momentum of a Lion of mass 130-kg and moving at a speed of 22.3 m/s [W]
Sunny_sXe [5.5K]

Answer:

<h2>289.9 kg.m/s</h2>

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 130 × 22.3

We have the final answer as

<h3>289.9 kg.m/s</h3>

Hope this helps you

3 0
3 years ago
A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s
aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is U_{s} = 0

And, initial potential gravitational potential energy of the rod is U_{g} = \frac{mgL}{2}.

It is given that,

       mass of the bar = 0.795 kg

            g = 9.8 m/s^{2}

           L = length of the rod = 0.2 m

Initial total energy T = \frac{mgL}{2}

Now, when the rod is in horizontal position then final total energy will be as follows.

            T = \frac{1}{2}kx^{2} + I \omega^{2}

where,    I = moment of inertia of the rod about the end = \frac{mL^{2}}{3}

Also,    \omega = \frac{\nu}{L}

where,    \nu = speed of the tip of the rod

              x = spring extension

The initial unstrained length is x_{o} = 0.1 m

Therefore, final length will be calculated as follows.

              x' = \sqrt{(0.2)^{2} + (0.1)^{2}} m

Then,  x = x' - x_{o}

          x = \sqrt{(0.2)^{2} + (0.1)^{2}} m - 0.1 m

             = 0.1236 m

       k = 25 N/m

So, according to the law of conservation of energy

       \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}

      \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

Putting the given values into the above formula as follows.

   \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

  \frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}

          v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0
3 years ago
In what type of change is matter not destroyed
Margaret [11]

There are two types of change in matter: physical change and chemical change. ... This is called the Law of Conservation of Matter. It states that matter can never be created or destroyed, only changed and rearranged.

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The answer is c or h
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Which are precautions to take when working with heat and fire?
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Always wear protected face gear and body gear, make sure you have an adult present and a fire extinguisher
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