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Juliette [100K]
3 years ago
14

A block of mass m=9.0 kg and speed V and is behind a block of mass M= 27 kg and speed of .50 m/s. The surface is frictionless, a

nd the blocks suddenly collide and couple. After the collision, the blocks have a common speed of .90 m/s. what was the speed v the lighter block before the collision'
Physics
1 answer:
sammy [17]3 years ago
5 0

Answer:

2.06 m/s

Explanation:

From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.

Momentum before collision

Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5

Momentum after collision

The momentum after collision will be given by (9+27)*0.9=32.4

Relating the two then 9v+13.5=32.4

9v=18.5

V=2.055555555555555555555555555555555555555 m/s

Rounded off, v is approximately 2.06 m/s

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Electrical metallic tubing would be used in an electrical installation because it
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Jasmine is diving off a 3-meter springboard. her height in meters above the water when she is x meters horizontally from the end
olganol [36]

Answer:

5.25 m

Explanation:

Given;

The height equation h;

h=-x^2+3x+3

Where;

h = the height above water

x = horizontal distance from the end of the board

The maximum height is at h' = 0, when change in h with respect to change in x is equal to zero.

differentiating the equation h.

dh/dx = h' = -2x + 3 = 0

Solving for x;

2x = 3

x = 3/2

Substituting into the function h;

h max = -x^2+3x+3

h max = -(3/2)^2 + 3(3/2) +3 = -9/4 +9/2 +3 = 9/4 + 3 =

h max = 21/4 = 5.25 m

8 0
3 years ago
What is the volume of this bubble when it reaches the surface?
steposvetlana [31]

Answer:

Volume will be 15 mL. Solution:- If we look at the given information then it is Boyle's law as the temperature is constant and the volume changes inversely as the pressure changes. So, the volume of the air bubble at the surface will be 15 mL.

8 0
3 years ago
2.0 mol of monatomic gas A initially has 5000 J of thermal energy. It interacts with 3.0 mol of monatomic gas B, which initially
Naddika [18.5K]

Answer:

 E_particle = 1,129 10⁻²⁰ J / particle

  T= 817.5 K

Explanation:

Energy is a scalar quantity so it is additive, let's look for the total energy of each gas

Gas a

         E_a = 2 5000 = 10000 J

Gas b

         E_b = 3 8000 = 24000 J

When the total system energy is mixed it is

          E_total = E_a + E_b

          E_total = 10000 + 24000 = 34000

The total mass is

           M = m_a + m_b

           M = 2 +3 = 5

The average energy among the entire mass is

           E_averge = E_total / M

            E_averago = 34000/5

            E_average = 6800 J

One mole of matter has Avogadro's number of atoms 6,022 10²³ particles

Therefore, each particle has an energy of

                E_particle = E_averag / 6.022 10²³ = 6800 /6.022 10²³

                E_particle = 1,129 10⁻²⁰ J / particle

For  find the temperature let's use equation

               E = kT

               T = E / k

     

               T = 1,129 10⁻²⁰ / 1,381 10⁻²³

               T = 8.175 102 K

               T= 817.5 K

5 0
4 years ago
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