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bazaltina [42]
3 years ago
9

What aspect of bouncing balls relates to matter

Physics
1 answer:
Lynna [10]3 years ago
8 0

Answer: An aspect of the event of various types of balls bouncing off the same floor, being matter is that all the balls consist of matter. They all occupy space and have a form of energy when moved by a force, such as a person. And for energy, like I just said, when they bounce they create energy as they bounce up and down, so if the ball were to hit some other object, it would have an impact on the still object.

The combination of the material properties of a ball (surface textures, actual materials, amount of air, hardness/ softness, and so on) affects the height of its bounce.

Hope this helps.......... Stay safe and have a Merry Christmas!!!!!!!!!! :D

Explanation:

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An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the
Andrei [34K]

Answer:

a) 2nd case rate of rotation gives the greater speed for the ball

b) 1534.98 m/s^2

c) 1515.04 m/s^2

Explanation:

(a) v = ωR

when R = 0.60, ω = 8.05×2π

v = 0.60×8.05×2π = 30.34 m/s

Now in 2nd case

when R = 0.90, ω = 6.53×2π

v = 0.90×6.53×2π = 36.92 m/s

6.35 rev/s gives greater speed for the ball.

(b) a = ω^2 R = (8.05×2π)^2 )(0.60) = 1534.98 m/s^2

(c) a = ω^2 R = (6.53×2π)^2 )(0.90) = 1515.05 m/s^2

7 0
3 years ago
You decide to travel by car for your holiday visits this year. You leave early in the morning to avoid congestion on the roads.
podryga [215]

Answer:

a) d = 182.08 miles

b) \overline{v} = 54.5 mph      

Explanation:

a) The distance can be found as follows:

d_{T} = d_{1} + d_{2} + d_{3}

d_{T} = v_{1}*t_{1} + v_{2}*t_{2} + v_{3}*t_{3}

d_{T} = 67.4 mph*1.70 h + 0*23.4 min + 54.0 mph*1.25 h = 182.08 miles = 292.9 km

b) The average speed can be calculated using the following equation:

\overline{v} = \frac{d_{f} - d_{i}}{t_{f} - t_{i}}

Where "f" is for final and "i" for initial

\overline{v} = \frac{182.08 miles - 0 miles}{(1.70 h + 23.4 min*\frac{1 h}{60 min} + 1.25 h) - 0 h} = 54.5 mph                

I hope it helps you!

5 0
4 years ago
State Newton's second law of<br>motion and write down<br>three examples to<br>justity it.​
kakasveta [241]

Answer:

The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

f = m \times a

f= force

m=mass

a=acceleration

Explanation:

examples:

riding your bicycle

•your bicycle is the mass, your leg pushing in pedals of your bicycle is the force

pushing a box

•the box is the mass, you are pushing the box

setting a pencil down in a table

•the pencil is the mass, you are puting the pencil down

3 0
3 years ago
How do i calculate this?
Lesechka [4]

CAR 1

Momentum = Mass/Velocity

M = 2100/20

M = 105 m/s^2

CAR 2

Momentum = Mass/Velocity

M = 2100/30

M = 70 m/s^2

8 0
3 years ago
Air enters the diffuser of a jet engine operating at steady state at 18 kPa, 216 K and a velocity of 265 m/s, all data correspon
earnstyle [38]

Answer:

45.44m/s

Explanation:

To solve this problem it is necessary to go back to the concepts related to the first law of thermodynamics,

in which it deepens on the conservation of the Energy.

The first law of Thermodynamics is given by the equation:

0 = \dot{Q}-\dot{W}+\dot{m}(h_1-h_2)+\dot{m}(\frac{V_1^2-V^2_2}{2})+\dot{m}g(z_1-z_2)

Where,

\dot{Q}= Heat transfer

\dot{W} =Work

\dot{m} =Flow mass

V_i =Velocity

h_i = Specific Enthalpy

g =Gravity

z_i =Height

From this equation we have that there is not Heat transfer, Work and changes in Height. Then,

Then our equation would be,

0=\dot{m}(h_1-h_2)+\dot{m}(\frac{V_1^2-V^2_2}{2})

Solving for V_2,

V_2 = \sqrt{V_1^2+2(h_1-h_2)}

From the tables of ideal gas (air) at 216K we have,

h_1 = 209.97+(219.97-209.97)(\frac{216-210}{220-210})

h_1 = 215.97kJ/kg

From the tables at 250K, we have that

h_2 = 250.05kJ/kg

The velocity was previously given, then

V_1 =265m/s

Replacing in the equation:

V_2 = \sqrt{265^2+2(215.97-250.05)*10^3}

V_2 = 45.44m/s

Therefore the velocity of the air at the diffuser exit is 45.44m/s

3 0
4 years ago
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