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bazaltina [42]
3 years ago
7

A billiard ball of mass 0.28 kg hits a second, identical ball at a speed of 7.2 m/s and comes to rest as the second ball flies o

ff. The collision takes 250 μs. What is the average force on the first ball? The second ball?
Physics
1 answer:
Strike441 [17]3 years ago
3 0

Answer:

(a) -8064 N

(b) 8064 N

Explanation:

(a)

From Newton’s law of motion, Force, F=ma where m is mass and a is acceleration.

Since acceleration is the rate of change of velocity per unit time, then where v is velocity and the subscripts f and I denote final and initial

For the first ball, the mass is 0.28 Kg, final velocity is zero since it finally comes to rest, t is 0.00025 s and initial velocity is given as 7.2 s. Substituting these values we obtain

F=0.28\times \frac {0-7.2}{0.00025}=-8064 N

(b)

For the second ball, the mass is also 0.28 Kg but its initial velocity is taken as zero, the final velocity of the second ball will be equal to the initial velocity of the second ball, that is 7.2 m/s and the time is also same, 0.00025 s. By substitution

F=0.28\times \frac {7.2-0}{0.00025}=8064 N

Here, we prove that action and reaction are equal and opposite

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Answer:

Friction is a force that holds back the movement of a sliding object.

Explanation:

The two types of friction: Static friction and Kinetic friction.  Static friction operates between two surfaces that aren't moving relative to each other, while kinetic friction acts between objects in motion.

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Advantage:  

Apparent solar time gives the exact location of sun in the sky according to your precise location.

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As the apparent solar time changes with the change in longitude. It is very difficult to track these changes in longitude. Hence, it is almost impossible to make plan for events.

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3 years ago
Resistivity of metallic wire
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6 0
3 years ago
1. A pumpkin with a mass of 2 kg accelerates 2 m/s/s when an unknown force is applied to it. What is the amount of the force?
hram777 [196]

Answer:

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6 0
3 years ago
A(n) 1700 kg car is moving along a level road at 21 m/s. The driver accelerates, and in the next 10 s the engine provides 22000
allochka39001 [22]

The final speed of the car at the given conditions is 30.1 m/s.

The given parameters:

  • <em>Mass of the car, m = 1700 kg</em>
  • <em>Velocity of the car, v = 21 m/s</em>
  • <em>Time of motion, t = 10 s</em>
  • <em>Additional energy provided by the engine, E₁ = 22,000 J</em>
  • <em>Energy used in overcoming friction, E₂ = 3,666.67 J</em>

The change in the energy applied to the car is calculated as;

\Delta E = E_1 - E_2\\\\\Delta E = 22,000 \ J \ - \ 3,666.67 \ J\\\\\Delta  E = 18,333.33 \ J

The final speed of the car is calculated as follows;

\Delta E = \frac{1}{2} m(v_f^2 - v_0^2)\\\\v_f^2 - v_0^2 = \frac{2\Delta E}{m} \\\\v_f^2  = \frac{2\Delta E}{m} + v_0^2\\\\v_f = \sqrt{ \frac{2\Delta E}{m} + v_0^2} \\\\v_f = \sqrt{ \frac{2\times 18,333.4}{1700} + (21)^2} \\\\v_f = 30.1 \ m/s

Thus, the final speed of the car at the given conditions is 30.1 m/s.

Learn more about change in kinetic energy here: brainly.com/question/6480366

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3 years ago
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