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Tamiku [17]
1 year ago
5

Two objects, m1 = 0.6 kg and m2 = 4.4 kg undergo a one-dimensional head-on collision

Physics
2 answers:
Temka [501]1 year ago
6 0

Explanation:

Hello !

look at the attachment above ☝️ and if you have any questions your welcome.

Sidana [21]1 year ago
4 0

a) The velocity after the collision.is 11.456 m/s.

b) The kinetic energy lost due to the collision is 44.564 J.

<h3>What is conservation of momentum principle?</h3>

When two bodies of different masses move together each other and have head on collision, they travel to same or different direction after collision.

The external force is not acting here, so the initial momentum is equal to the final momentum. For inelastic collision, final velocity is the common velocity for both the bodies.

m₁u₁ +m₂u₂ =(m₁ +m₂) v

Given are the two objects, m1 = 0.6 kg and m2 = 4.4 kg undergo a one-dimensional head-on collision. Their initial velocities along the one-dimension path are vi1 = 32.4 m/s [right] and vi2 = 8.6 m/s [left].

(a) Substitute the values, then the final velocity will be

0.6 x32.4 +4.4 x 8.6 = (0.6+4.4)v

v = 11.456 m/s

Thus, the velocity after collision is 11.456 m/s.

(b) Kinetic energy lost due to collision will be the difference between the kinetic energy before and after collision.

= [1/2m₁u₁² +1/2m₂u₂² ] - [1/2(m₁ +m₂) v²]

Substitute the value, we have

= [1/2 x 0.6 x32.4² + 1/2 x4.4 x 8.6²] - [1/2 x(0.6+4.4)11.456²]

= 44.564 J

Thus, the kinetic energy lost due to the collision is 44.564 J.

Learn more about conservation of momentum principle

brainly.com/question/14033058

#SPJ2

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Inductors in parallel. Two inductors L1 = 1.31 H and L2 = 2.24 H are connected in parallel and separated by a large distance so
mote1985 [20]

Answer:

a) 0.83H

b) 3.22/N Henry

Explanation:

Given two inductors L1 = 1.31 H and L2 = 2.24 H connected in parallel, their equivalent inductance derivative is similar to that of resistance in parallel.

Derivation:

If the voltage across an inductor

VL = IXL

I is the current

XL is the inductive reactance

XL = 2πfL

VL = I(2πfL)

L is the inductance.

From the formula, I = V/2πfL

Given two inductors in parallel, different current will flow through them.

I1 = V/2πfL1 (current in L1)

I2 = V/2πfL2 (current in L2)

Total current I = I1+I2

I = V/2πfL1 + V/2πfL2

I = V/2πf{1/L1+1/L2}

V/2πfL = V/2πf{1/L1+1/L2}

1/L = 1/L1+1/L2 (equivalent inductance in parallel)

Given L1 = 1.31 H and L2 = 2.24

1/L = 1/1.31 + 1/2.24

1/L = 0.763 + 0.446

1/L = 1.209

L = 1/1.209

L = 0.83H

The equivalent inductance is 0.83H

b) Given similar inductors L = 3.22H in parallel, the equivalent inductance will be:

1/L = 1/3.22+1/3.22+1/3.22+1/3.22+1/3.22

+1/3.22+1/3.22+1/3.22+1/3.22+1/3.22

1/L = 10/3.22 (since that all have the same denominator)

L = 3.22/10

If N = 10, the generalization of 10 similar inductors in parallel will be:

L = 3.22/N Henry

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