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Rudiy27
1 year ago
13

Question 2

Physics
1 answer:
ArbitrLikvidat [17]1 year ago
5 0

a = Δv + xt^2 is not  a kinematic equation.

<h3>What is a kinematic equation?</h3>

Kinematic equations are those equations that equations that could be used to describe the motion of a body without involving the force that makes the object to move. In considering the kinematic equations, we have to consider the following;

  • Initial velocity
  • Final velocity
  • position
  • acceleration

Thus, looking at the kinematic equations as shown, we can see that the equation; a = Δv + xt^2 is not  a kinematic equation.

Learn more about kinematics:brainly.com/question/7590442

#SPJ1

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What is absolute zero? What is the temperature of absolute zero on the Kelvin and Celsius scales?
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Answer:

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Explanation:

Absolute zero :

 When the entropy and enthalpy of the ideal system reach at the minimum value then the temperature at that condition is known as absolute zero condition.

Absolute temperature is the minimum temperature in the temperature scale.The value of absolute zero is 0 K.

We know that

\dfrac{C-0}{100}=\dfrac{K-273}{100}=\dfrac{F-32}{180}

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When  K = 0

\dfrac{C-0}{100}=\dfrac{K-273}{100}

\dfrac{C-0}{100}=\dfrac{0-273}{100}

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On the right side, there are 4 Fe atoms and 6 O atoms.

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A rocket is fired vertically upward. At the instant it reaches some altitude with a speed of 100 m/s. It explodes into three fra
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Answer:

The magnitude of the velocity of the third fragment  is 212.13 m/s.

Explanation:

Given that,

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Vertical speed of second fragment = 150 m/s

We need to calculate the vertical speed of third fragment

Using conservation of momentum

P_{iy}=P_{fy}

mv=\dfrac{m}{3}{v_{1}}+\dfrac{m}{3}v_{2}+\dfrac{m}{3}v_{3}

Put the value into the formula

m\times100=\dfrac{m}{3}\times150+\dfrac{m}{3}\times0+\dfrac{m}{3}\times v_{y}

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v_{y}=150\ m/s

We need to calculate the horizontal speed of third fragment

Using conservation of momentum

P_{iy}=P_{fy}

mv=\dfrac{m}{3}{v_{1}}+\dfrac{m}{3}v_{2}+\dfrac{m}{3}v_{3}

Put the value into the formula

m\times0=\dfrac{m}{3}\times0+\dfrac{m}{3}\times150+\dfrac{m}{3}v_{x}

\dfrac{1}{3}v_{x}=-50

v_{x}=-150\ m/s

We need to calculate the magnitude of the velocity of the third fragment

v=\sqrt{v_{x}^2+v_{y}^2}

Put the value into the formula

v=\sqrt{(-150)^2+(150)^2}

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Hence, The magnitude of the velocity of the third fragment  is 212.13 m/s.

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