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xeze [42]
3 years ago
14

Object A has of mass 7.20 kilograms, and object B has a mass of 5.75 kilograms. The two objects move along a straight line towar

d each other with velocities +2.00 meters/second and -1.30 meters/second respectively. What is the total kinetic energy of the objects after the collision, if the collision is perfectly elastic?
Physics
2 answers:
cestrela7 [59]3 years ago
8 0
Two things:
1) All collisions conserve momentum
2) Elastic collisions also conserve kinetic energy.

So basically the kinetic energy before the collision is the same as the kinetic energy after the collision.

KE = ½mv²

For object A:
KE = ½(7.20 kg)(2.00 m/s)² = 14.4 J

For object B:
KE = ½(5.75 kg)(-1.30 m/s)² = 4.86 J

total kinetic energy = 14.4 J + 4.86 J = 19.3 J
Genrish500 [490]3 years ago
7 0
The equation for Kinetic Energy in an Elastic Collision is this:
1/2(m₁v₁) + 1/2(m₂v₂) = 1/2(7.20kg)(2.00m/s) + 1/2(5.75kg)(-1.30m/s) =
3.4625m/s

I think....technically the equation above equates to the same thing except measuring the final velocity of the given masses, however you were not given that in the equation, making the first half of the full equation the only viable solution.

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12.51 A parallel RLC circuit, which is driven by a variable frequency 2-A current source, has the following values: R = 1 kΩ, L
Anastaziya [24]

Answer:

BW = 100 rad/s

wlow = 452.49 rad/s

whigh = 552.49 rad/s

V(jwlow) =1414.21 < 45°V

V(jwhigh) =1414.21 <-45°V

Explanation:

To calculate bandwidth we have formula

BW = 1/RC

BW = 1/ 1000x10x10^¯6

BW = 100 rad/s

We will first calculate resonant frequency and quality factor for half power frequencies.

For resonant frequency

wo = 1/(SQRT LC)

wo = 1/SQRT 400×10¯³ × 10×10^¯6

wo = 500 rad/s

For Quality

Q = wo / BW

Q = 500/100

Q = 5

wlow = wo [-1/2Q+ SQRT (1/2Q)² + 1]

wlow = 500 [-1/2×5 + SQRT (1/2×5)² + 1]

wlow = 452.49 rad/s

whigh = wo [1/2Q+ SQRT (1/2Q)² + 1]

whigh = 500 [1/2×5 + SQRT (1/2×5)² + 1]

whigh = 552.49 rad/s

We will start with admittance at lower half power frequency

Y(jwlow) = (1/R) + (1/jwlow L) + (jwlow C)

Y(jwlow) = (1/1000) + (1/j×452.49×400×10¯³) + (j×452.49×10×10^¯6)

Y(jwlow) = 0.001 - j5.525×10¯³ + j4.525×10¯³

Y(jwlow) = (1-j).10¯³ S

Voltage across the network is calculated by ohm's law

V(jwlow) = I/Y(jwlow)

V(jwlow) = 2/(1-j).10¯³

V(jwlow) = 1414.2 < 45°V

Now we will calculate the admittance at higher half power frequency

Y(jwhigh) = (1/R) + (1/jwhigh L) + (jwhigh C)

Y(jwhigh) = (1/1000) + (1/j×552.49×400×10¯³) + (j×552.49×10×10^¯6)

Y(jwhigh) = 0.001 - j4.525×10¯³ + j5.525×10¯³

Y(jwhigh) = (1+j).10¯³ S

Voltage across network will be calculated by ohm's law

V(jwhigh) = I/Y(jwhigh)

V(jwhigh) = 2/(1+j).10¯³

V(jwhigh) = 1414.2 < - 45°V

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