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Basile [38]
3 years ago
14

1. What is the Potential Energy of a 6 kg object that is 10 m above the ground?

Mathematics
1 answer:
andriy [413]3 years ago
5 0
PE = mgh
m = 6 kg, g = 9.8 m/s^2, h = 10m
6 * 9.8 * 10 = 588 J
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Which equations are true?
Vlad [161]

Answer:

Step-by-step explanation:

1)4.5m+4.7=20m+47

4.5m-20m=47-4.7

-15.5m=42.2

m=-2.72

2)14+21w=7(2+3w)

  41+21w=14+21w

   21w-21w=14-41

3)49r+35=7(7r+35)

  49r+35=49r+245

 49r-49r=245-35

4)9(8h-3)=72h-27

  72h-27=72h-27

   72h-72h=-27+27

5)5.2+5.3t=10+15t

   5.3t-15t=10-5.2

    -9.7t=4.8

      t=4.8/-9.7

      t=0.49

6)6f+3.111=18f-33

  6f-18f=-33-3.111

  -12f=-36.111

     f=-36.111/-12

     f=3.01

7 0
3 years ago
Find the surface area of this triangular prism. Be sure to include the correct unit in your answer.
BartSMP [9]

Answer:

905 is the perimeter

and the triangular prism is 46

Step-by-step explanation:

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4 0
3 years ago
X y<br> –<br> 10 700<br> –<br> 9 567<br> –<br> 8 448<br> –<br> 7 343<br> –<br> 6 252
JulsSmile [24]

Answer:

I'm not answering your question but I wanted to tell y'all to have a good day!<3

Step-by-step explanation:

6 0
2 years ago
Find the equation of the line passing through the points (1,5) and (3,9)​
Arada [10]

Answer:

  • \boxed{\sf{2}}

Step-by-step explanation:

Use the slope formula.

\underline{\text{SLOPE:}}

\Longrightarrow: \sf{\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{Rise}{Run} }

\sf{y_2=9}\\\\\\\sf{y_1=5}\\\\\\\sf{x_2=3}\\\\\\\sf{x_1=1}

\sf{\dfrac{9-5}{3-1} }

Solve.

\sf{\dfrac{9-5}{3-1}=\dfrac{4}{2}=\boxed{\sf{2}}

  • <u>Therefore, the slope is 2, which is our answer.</u>

I hope this helps. Let me know if you have any questions.

8 0
1 year ago
y = c1 cos(5x) + c2 sin(5x) is a two-parameter family of solutions of the second-order DE y'' + 25y = 0. If possible, find a sol
TEA [102]

Answer:

y = 2cos5x-9/5sin5x

Step-by-step explanation:

Given the solution to the differential equation y'' + 25y = 0 to be

y = c1 cos(5x) + c2 sin(5x). In order to find the solution to the differential equation given the boundary conditions y(0) = 1, y'(π) = 9, we need to first get the constant c1 and c2 and substitute the values back into the original solution.

According to the boundary condition y(0) = 2, it means when x = 0, y = 2

On substituting;

2 = c1cos(5(0)) + c2sin(5(0))

2 = c1cos0+c2sin0

2 = c1 + 0

c1 = 2

Substituting the other boundary condition y'(π) = 9, to do that we need to first get the first differential of y(x) i.e y'(x). Given

y(x) = c1cos5x + c2sin5x

y'(x) = -5c1sin5x + 5c2cos5x

If y'(π) = 9, this means when x = π, y'(x) = 9

On substituting;

9 = -5c1sin5π + 5c2cos5π

9 = -5c1(0) + 5c2(-1)

9 = 0-5c2

-5c2 = 9

c2 = -9/5

Substituting c1 = 2 and c2 = -9/5 into the solution to the general differential equation

y = c1 cos(5x) + c2 sin(5x) will give

y = 2cos5x-9/5sin5x

The final expression gives the required solution to the differential equation.

3 0
3 years ago
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