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oksano4ka [1.4K]
3 years ago
9

How would you solve for x I cant remember right now 4x+6x=9x-10

Physics
2 answers:
-Dominant- [34]3 years ago
8 0
Combine all of the x's on one side of the equation and then finish the problem!
stepan [7]3 years ago
4 0
So you are first going to want to combine the like terms (in this case, 4x + 6x = 10x) so that the equation is 10x = 9x-10. Next, subtract 9x from both sides. 
10x = 9x - 10
-9x    -9x 
-------------------
 x    =   -10

Sometimes there are more steps, but in general, as long as you do the addition and subtraction before any multiplication, division, or exponential work, you will not have many problems.
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M/s
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Answer:

a. Final velocity, V = 2.179 m/s.  

b. Final velocity, V = 7.071 m/s.

Explanation:

<u>Given the following data;</u>

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a. To find the velocity of the boat after it has traveled 4.75 m

Since it started from rest, initial velocity is equal to 0m/s.

Now, we would use the third equation of motion to find the final velocity.

V^{2} = U^{2} + 2aS

Where;

  • V represents the final velocity measured in meter per seconds.
  • U represents the initial velocity measured in meter per seconds.
  • a represents acceleration measured in meters per seconds square.
  • S represents the displacement measured in meters.

Substituting into the equation, we have;

V^{2} = 0^{2} + 2*0.500*4.75

V^{2} = 4.75

Taking the square root, we have;

V^{2} = \sqrt {4.75}

<em>Final velocity, V = 2.179 m/s.</em>

b. To find the velocity if the boat has traveled 50 m.

V^{2} = 0^{2} + 2*0.500*50

V^{2} = 50

Taking the square root, we have;

V^{2} = \sqrt {50}

<em>Final velocity, V = 7.071 m/s.</em>

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A mass m attached to a horizontal massless spring with spring constant k, is set into simple harmonic motion. its maximum displa
Lesechka [4]
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E_i = U_i+K= \frac{1}{2}ka^2

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U_f = 0
while the mass is moving at speed v, and therefore the kinetic energy is
K_f =  \frac{1}{2} mv^2
And the total energy is
E_f = U_f + K_f =  \frac{1}{2} mv^2

For the law of conservation of energy, the total energy must be conserved, therefore E_i = E_f. So we  can write
\frac{1}{2} ka^2 =  \frac{1}{2}mv^2
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