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miss Akunina [59]
1 year ago
5

in an inelastic collision, kinetic energy is not conserved. where does it go? (select all that apply)

Physics
1 answer:
Rama09 [41]1 year ago
5 0

In an elastic collision, although total energy of a system is always conserved, kinetic energy carried by the moving objects is not always conserved. It is lost in the environment.

In an inelastic collision, energy is lost in the environment and transferred into other forms such as heat. When total kinetic energy is not conserved, then that collision is called as an inelastic collision.

Although momentum is conserved in inelastic collisions but it is difficult to track the kinetic energy through the collision since some of it is converted into other forms of energy.

An inelastic collision in a ballistic pendulum. Another example of an inelastic collision is a ballistic pendulum or dropped ball of clay. A dropped ball of clay doesn't rebound instead it loses kinetic energy through deformation on hitting the ground and changes shape.

To know more about inelastic collisions, refer

brainly.com/question/24434011

#SPJ4

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0.8 meters.

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Outside the Solar System is there any gravitational pull from the Sun?
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Which technique is best for manual in-line stabilization of a person floating faceup on the surface?
zmey [24]

Answer:

vise grip

Explanation:

Manual in-line stabilization (MILS) of the cervical spine is a type of airway management when dealing with  patients in traumatic condition ..it is a means that is performed by grasping the mastoid process of the patient, so as to prevent the movement of the cervical column during intubation of the trachea

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3 years ago
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In 1970, a rocket powered car called Blue Flame achieved a maximum speed of 1.00(10 km/h (278m/s).Suppose the magnitude of the c
sammy [17]

Answer:

Distance traveled during this acceleration will be 6950 m

Explanation:

Wear have given maximum speed tat will be equal to final speed of the car        v = 278 m/sec

Constant acceleration a=5.56m/sec^2

As the car starts initially starts from rest so initial velocity of the car u = 0 m/sec

From third equation of motion v^2=u^2+2as

Putting all values in equation

278^2=0^2+2\times 5.56\times s

s = 6950 m

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3 0
4 years ago
Point charges q1 = 14 µC and q2 = −60 µC are fixed at r1 = (5.0î − 4.0ĵ) m and r2 = (9.0î + 7.5ĵ) m. What is the force (in N) of
Lostsunrise [7]

Answer:

The force on q₁ due to q₂ is (0.00973i + 0.02798j) N

Explanation:

F₂₁ = \frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|}

Where;

F₂₁ is the vector force on q₁ due to q₂

K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²

r₂₁ is the unit vector

|r₂₁| is the magnitude of the unit vector

|q₁| is the absolute charge on point charge one

|q₂| is the absolute charge on point charge two

r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)

|r₂₁| = \sqrt{(4^2)+(11.5^2)} = \sqrt{148.25}

(|r₂₁|)² = 148.25

F_2_1=\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|} = \frac{8.99X10^9(14X10^{-6})(60X10^{-6})}{148.25}.\frac{(4i + 11.5j)}{\sqrt{148.25} }

      = 0.050938(0.19107i + 0.54933j) N

      = (0.00973i + 0.02798j) N

Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N

7 0
4 years ago
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