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Paraphin [41]
2 years ago
15

Do as physics instructor fred cauthen does and place a tennis ball close to and above the top of a basketball. drop the balls to

gether. if their vertical alignment nicely remains as they fall to the floor, you'll see that the tennis ball bounces unusually high. can you reconcile this with energy conservation?
Physics
1 answer:
Ksivusya [100]2 years ago
6 0

Answer:

after shock

creating a system for the conservation of the energy of the basketball ball and creating a system for the tennis ball only, the conservation of energy should be applied to each system independently

Explanation:

When the two balls fall they acquire the same speed since they are accelerated by the same force, their weight and the acceleration of the acceleration of gravity. When reaching the floor the mechanical energy of the system is conserved.

Upon reaching the floor, the first ball (basketball) collides with the floor, this process is very fast, at the end of the process the basketball comes out with a velicad up and collides with the much lighter tennis ball that is still descending .

we assume that the shocks are elastic, when solving the momentary and kinetic energy findings, we find the velocities after each shock

     

In this clash the tennis ball acquires a high kinetic speed with an upward direction that makes a very high height high. Again this shock is very fast and the tennis ball almost does not move.

Here we must separate the system, creating a system for the conservation of the energy of the basketball ball and another system for the tennis ball only, the conservation of energy should be applied to each system independently

Em₀ =K = 1/2 m v²

                    Em_{f} = U = m g h

As in the elastic shock the final speed of the tennis ball is approximately 2 vo, we can calculate the maximum height

                 m g h = 1/2 m (2v₀)²

                 h = 2 v₀²/g

To reconcile this with the conservation of energy we must calculate the energy for the tennis ball at two points, the first when the crash with the tennis ball ends and at the end point at its maximum height.

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An 8.0-kg history textbook is placed on a 1.25-m high desk. How large is the gravitational potential energy of the textbook-Eart
denpristay [2]
<h2>Answer</h2>

Gravitational potential energy of the textbook-Earth system relative to the desk = 98 N/m

<h2>Explanation</h2>

Given that,

Mass of history book = 8kg

height of book from ground = 1.25m

acceleration due to gravity = 9.8m/s²

<h3>gravitational potential energy formula</h3><h2>GPE = Fg⋅h</h2><h2>GPE = mgh</h2>

where ,

m = mass of object(book)

g = acceleration due to gravity

h = the altitude of the object.

GPE = 8(9.8)(1.25)

       = 98 N/m

So, gravitational potential energy of the textbook-Earth system relative to the desk = 98 N/m

5 0
2 years ago
Read 2 more answers
If radio waves are used to communicate with an alien spaceship approaching Earth at 10% of the speed of light c, the aliens woul
Brilliant_brown [7]

Answer:

3×10^7 m/s or 0.10c (e)

Explanation: If the actual value of the speed of light were to be put into consideration.

Given that the speed of light is c = 3.0×10^8m/s

The alien spaceship is approaching at the rate of 10% of the speed of light.

10% of 3.0×10^8m/s

10/100 × 3.0×10^8m/s

0.1 ×3.0×10^8m/s

3×10^7 m/s. Which is the same thing as 0.1 of c = 0.1×c

7 0
2 years ago
Suppose two astronauts on a spacewalk are floating motionless in space, 3.0 m apart. Astronaut B tosses a 15.0 kg IMAX camera to
marta [7]

Answer:

\frac{ 112.5}{15+m_{A}}=v_{f}

(we need the mass of the astronaut A)

Explanation:

We can solve this by using the conservation law of the linear momentum P. First we need to represent every mass as a particle. Also we can simplify this system of particles by considering only the astronaut A with an initial speed v_{iA} of 0 m/s and a mass m_{A} and the IMAX camera with an initial speed v_{ic} of 7.5 m/s and a mass m_{c} of 15.0 kg.

The law of conservation says that the linear momentum P (the sum of the products between all masses and its speeds) is constant in time. The equation for this is:

P_{i}=p_{ic}+p_{iA}\\P_{i}=m_{c}v_{ic}+m_{A} v_{iA}\\P_{i}=15*7.5 + m_{A}*0\\P_{i}=112.5 \frac{kg.m}{s}

By the law of conservation we know that P_{i} =P_{f}

For P_{f} (final linear momentum) we need to treat the collision as a plastic one (the two particles stick together after the encounter).

So:

P_{i} =P_{f}=112.5\\

112.5=(m_{c}+m_{A})v_{f}\\\frac{ 112.5}{m_{c}+m_{A}}=v_{f}\\\frac{ 112.5}{15+m_{A}}=v_{f}

3 0
2 years ago
Which type of listening response includes the use of head nods, facial expressions, and short utterances such as "uh-huh" that s
BARSIC [14]
This type of listening response is called back-channel signal. This allows the speaker to know that the listener is attentive or willing to engage a conversation between them. It is shown through short utterances, facial expressions, head nods and others. 
4 0
2 years ago
An archer draws her bow and stores 34.8 J of elastic potential energy in the bow. She releases the 63 g arrow, giving it an init
elena-14-01-66 [18.8K]

Answer:

Approximately 71\%.

Explanation:

The formula for the kinetic energy \rm KE of an object is:

\displaystyle \mathrm{KE} = \frac{1}{2}\, m \cdot v^2,

where

  • m is the mass of that object, and
  • v is the speed of that object.

Important: Joule (\rm J) is the standard unit for energy. The formula for \rm KE requires two inputs: mass and speed. The standard unit of mass is \rm kg while the standard unit for speed is \rm m \cdot s^{-1}. If both inputs are in standard units, then the output (kinetic energy) will also be in the standard unit (that is: joules,

Convert the unit of the arrow's mass to standard unit:

m = 63\; \rm g = 0.063\; \rm kg.

Initial \rm KE of this arrow:

\begin{aligned}\mathrm{KE} &= \frac{1}{2} \, m \cdot v^2 \\ &= \frac{1}{2}\times 0.063\; \rm kg \times \left(\rm 28 \; m \cdot s^{-1}\right)^2 \\ &\approx 24.696\; \rm J\end{aligned}.

That's the same as the energy output of this bow. Hence, the efficiency of energy transfer will be:

\displaystyle \frac{24.696\; \rm J}{34.8\; \rm J} \times 100\% \approx 71\%.

8 0
2 years ago
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