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Tamiku [17]
2 years 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]2 years ago
6 0

Explanation:

Hello !

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

Sidana [21]2 years 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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A ladybug sits 14 cm from the center of a turntable spinning at 33.33 rpm. The Sun is shining horizontally through the window an
kolbaska11 [484]

Answer:

The maximum velocity is 0.489 m/s

Explanation:

Maximum velocity (v) = angular velocity (w) × radius (r)

w = 33.33 rpm = 33.33×0.1047 = 3.4897 rad/s

r = 14 cm = 14/100 = 0.14 m

v = 3.4897×0.14 = 0.489 m/s

3 0
3 years ago
If the power rating on a motor is 10W and 500J of energy is used to move a 20N object, how long is the motor used for?
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The power rating on the motor is the maximum power it's ABLE to deliver. It can run with LESS power output than the rating, but if you try to run it with MORE than it's rated, the motor will overheat and eventually burn out.

" 10 watts " means " 10 Joules of enrgy per second ".

If the motor is operated at its full maximum rated capacity, then

(500 joules) / (10 joules/sec) = <em>50 seconds</em>

4 0
4 years ago
Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton
solmaris [256]

Answer:

a

\lambda = 3.68 *10^{-36} \  m

b

\lambda_p = 1.28*10^{-14} \ m

Explanation:

From the question we are told that

   The mass of the person is  m =  180 \  kg

    The speed of the person is  v  =  1 \  m/s

    The energy of the proton is  E_ p =  5 MeV = 5 *10^{6} eV  = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \  J

Generally the de Broglie wavelength is mathematically represented as

      \lambda = \frac{h}{m * v }

Here  h is the Planck constant with the value

      h = 6.62607015 * 10^{-34} J \cdot s

So  

     \lambda = \frac{6.62607015 * 10^{-34}}{ 180  * 1  }

=> \lambda = 3.68 *10^{-36} \  m

Generally the energy of the proton is mathematically represented as

         E_p =  \frac{1}{2}  *   m_p  *  v^2_p

Here m_p  is the mass of proton with value  m_p  =  1.67 *10^{-27} \  kg

=>     8.0*10^{-13} =  \frac{1}{2}  *   1.67 *10^{-27}  *  v^2

=>   v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }

=>   v = 3.09529 *10^{7} \  m/s

So

        \lambda_p = \frac{h}{m_p * v_p }

so    \lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }

=>     \lambda_p = 1.28*10^{-14} \ m

     

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Oxygen... Hope this Warner helps
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